Automorphism Groups

Home , Automorphism, Automorphism group, Inner automorphism

Gerhardt Hinkle

Introduction Automorphisms Original problem Generalizations Prime difference Prime square Automorphism groups difference Possible differences Further research Gerhardt Hinkle

Missouri State University REU, 2013 I The set of automorphisms of G forms a group under function composition. The automorphism group of G is written Aut(G). I The inner automorphism group of G, written Inn(G), is −1 the group of automorphisms of the form fg (x) = gxg for a ﬁxed g ∈ G. I The center of G, written Z(G), comprises those elements of G that commute with every element of G. I Inn(G) =∼ G/Z(G) I G/Z(G) is the group of left cosets of Z(G) in G (i.e. sets of the form gZ(G) for some g ∈ G). It can also be thought of as taking G and setting every element of Z(G) equal to the identity element e. I |Inn(G)| divides |Aut(G)| and |G|.

Automorphism Automorphisms groups Gerhardt Hinkle I For a group G, an automorphism of G is a function f : G → G that is bijective and satisfies Introduction Automorphisms f (xy) = f (x)f (y) for all x, y ∈ G. Original problem Generalizations Prime difference Prime square difference Possible differences Further research I The inner automorphism group of G, written Inn(G), is −1 the group of automorphisms of the form fg (x) = gxg for a fixed g ∈ G. I The center of G, written Z(G), comprises those elements of G that commute with every element of G. I Inn(G) =∼ G/Z(G) I G/Z(G) is the group of left cosets of Z(G) in G (i.e. sets of the form gZ(G) for some g ∈ G). It can also be thought of as taking G and setting every element of Z(G) equal to the identity element e. I |Inn(G)| divides |Aut(G)| and |G|.

Automorphism Automorphisms groups Gerhardt Hinkle I For a group G, an automorphism of G is a function f : G → G that is bijective and satisfies Introduction Automorphisms f (xy) = f (x)f (y) for all x, y ∈ G. Original problem Generalizations I The set of automorphisms of G forms a group under Prime difference Prime square function composition. The automorphism group of G is difference written Aut(G). Possible differences Further research I The inner automorphism group of G, written Inn(G), is −1 the group of automorphisms of the form fg (x) = gxg for a fixed g ∈ G. I The center of G, written Z(G), comprises those elements of G that commute with every element of G. I Inn(G) =∼ G/Z(G) I G/Z(G) is the group of left cosets of Z(G) in G (i.e. sets of the form gZ(G) for some g ∈ G). It can also be thought of as taking G and setting every element of Z(G) equal to the identity element e. Automorphism Automorphisms groups Gerhardt Hinkle I For a group G, an automorphism of G is a function f : G → G that is bijective and satisfies Introduction Automorphisms f (xy) = f (x)f (y) for all x, y ∈ G. Original problem Generalizations I The set of automorphisms of G forms a group under Prime difference Prime square function composition. The automorphism group of G is difference written Aut(G). Possible differences Further research I The inner automorphism group of G, written Inn(G), is −1 the group of automorphisms of the form fg (x) = gxg for a fixed g ∈ G. I The center of G, written Z(G), comprises those elements of G that commute with every element of G. I Inn(G) =∼ G/Z(G) I G/Z(G) is the group of left cosets of Z(G) in G (i.e. sets of the form gZ(G) for some g ∈ G). It can also be thought of as taking G and setting every element of Z(G) equal to the identity element e. I |Inn(G)| divides |Aut(G)| and |G|. ∼ n I Zn = ha|a = 1i ∼ × × I Aut(Zn) = Zn , where Zn is the group of the integers relatively prime to n under multiplication mod n.

I |Aut(Zn)| = φ(n) I φ(n) is Euler’s totient function, which gives the number of positive integers less than or equal to n that are relatively prime to n.

I Fundamental theorem of ﬁnite abelian groups: Every ﬁnite abelian group is isomorphic to the direct product of some number of cyclic groups. ∼ I If gcd(m, n) = 1, then Zmn = Zm × Zn.

Automorphism Cyclic groups groups Gerhardt Hinkle I For a positive integer n, the cyclic group of order n, Introduction written Zn, is the group of order n generated by one Automorphisms element. It is isomorphic to the group of the integers Original problem Generalizations under addition mod n. Prime difference Prime square difference Possible differences Further research ∼ × × I Aut(Zn) = Zn , where Zn is the group of the integers relatively prime to n under multiplication mod n.

I |Aut(Zn)| = φ(n) I φ(n) is Euler’s totient function, which gives the number of positive integers less than or equal to n that are relatively prime to n.

I Fundamental theorem of ﬁnite abelian groups: Every ﬁnite abelian group is isomorphic to the direct product of some number of cyclic groups. ∼ I If gcd(m, n) = 1, then Zmn = Zm × Zn.

Automorphism Cyclic groups groups Gerhardt Hinkle I For a positive integer n, the cyclic group of order n, Introduction written Zn, is the group of order n generated by one Automorphisms element. It is isomorphic to the group of the integers Original problem Generalizations under addition mod n. Prime difference Prime square ∼ n difference I Zn = ha|a = 1i Possible differences Further research I |Aut(Zn)| = φ(n) I φ(n) is Euler’s totient function, which gives the number of positive integers less than or equal to n that are relatively prime to n.

I Fundamental theorem of ﬁnite abelian groups: Every ﬁnite abelian group is isomorphic to the direct product of some number of cyclic groups. ∼ I If gcd(m, n) = 1, then Zmn = Zm × Zn.

Automorphism Cyclic groups groups Gerhardt Hinkle I For a positive integer n, the cyclic group of order n, Introduction written Zn, is the group of order n generated by one Automorphisms element. It is isomorphic to the group of the integers Original problem Generalizations under addition mod n. Prime difference Prime square ∼ n difference I Zn = ha|a = 1i Possible differences ∼ × × Further research I Aut(Zn) = Zn , where Zn is the group of the integers relatively prime to n under multiplication mod n. I φ(n) is Euler’s totient function, which gives the number of positive integers less than or equal to n that are relatively prime to n.

I Fundamental theorem of ﬁnite abelian groups: Every ﬁnite abelian group is isomorphic to the direct product of some number of cyclic groups. ∼ I If gcd(m, n) = 1, then Zmn = Zm × Zn.

Automorphism Cyclic groups groups Gerhardt Hinkle I For a positive integer n, the cyclic group of order n, Introduction written Zn, is the group of order n generated by one Automorphisms element. It is isomorphic to the group of the integers Original problem Generalizations under addition mod n. Prime difference Prime square ∼ n difference I Zn = ha|a = 1i Possible differences ∼ × × Further research I Aut(Zn) = Zn , where Zn is the group of the integers relatively prime to n under multiplication mod n.

I |Aut(Zn)| = φ(n) I Fundamental theorem of ﬁnite abelian groups: Every ﬁnite abelian group is isomorphic to the direct product of some number of cyclic groups. ∼ I If gcd(m, n) = 1, then Zmn = Zm × Zn.

Automorphism Cyclic groups groups Gerhardt Hinkle I For a positive integer n, the cyclic group of order n, Introduction written Zn, is the group of order n generated by one Automorphisms element. It is isomorphic to the group of the integers Original problem Generalizations under addition mod n. Prime difference Prime square ∼ n difference I Zn = ha|a = 1i Possible differences ∼ × × Further research I Aut(Zn) = Zn , where Zn is the group of the integers relatively prime to n under multiplication mod n.

I |Aut(Zn)| = φ(n) I φ(n) is Euler’s totient function, which gives the number of positive integers less than or equal to n that are relatively prime to n. ∼ I If gcd(m, n) = 1, then Zmn = Zm × Zn.

Automorphism Cyclic groups groups Gerhardt Hinkle I For a positive integer n, the cyclic group of order n, Introduction written Zn, is the group of order n generated by one Automorphisms element. It is isomorphic to the group of the integers Original problem Generalizations under addition mod n. Prime difference Prime square ∼ n difference I Zn = ha|a = 1i Possible differences ∼ × × Further research I Aut(Zn) = Zn , where Zn is the group of the integers relatively prime to n under multiplication mod n.

I |Aut(Zn)| = φ(n) I φ(n) is Euler’s totient function, which gives the number of positive integers less than or equal to n that are relatively prime to n.

I Fundamental theorem of finite abelian groups: Every finite abelian group is isomorphic to the direct product of some number of cyclic groups. Automorphism Cyclic groups groups Gerhardt Hinkle I For a positive integer n, the cyclic group of order n, Introduction written Zn, is the group of order n generated by one Automorphisms element. It is isomorphic to the group of the integers Original problem Generalizations under addition mod n. Prime difference Prime square ∼ n difference I Zn = ha|a = 1i Possible differences ∼ × × Further research I Aut(Zn) = Zn , where Zn is the group of the integers relatively prime to n under multiplication mod n.

I |Aut(Zn)| = φ(n) I φ(n) is Euler’s totient function, which gives the number of positive integers less than or equal to n that are relatively prime to n.

I Fundamental theorem of ﬁnite abelian groups: Every ﬁnite abelian group is isomorphic to the direct product of some number of cyclic groups. ∼ I If gcd(m, n) = 1, then Zmn = Zm × Zn. ∼ I If n 6= 2, 6, then Aut(Sn) = Sn. Therefore, d(Sn) = 0 for all n 6= 2, 6.

Automorphism Original problem groups Gerhardt Hinkle

Automorphism Original problem (continued) groups Gerhardt Hinkle

Introduction Automorphisms Original problem I d(G) = |Aut(G)| − |G| = ±1 Generalizations Prime difference Prime square difference Possible differences Further research ∼ I G = Z a1 × Z a2 × ... × Z ak p1 p2 pk I Aut(G) ≥ Aut(Z a1 ) × Aut(Z a2 ) × ... × Aut(Z ak ) p1 p2 pk ai −1 I |Aut(Z ai )| = p (pi − 1) pi i I a1 = a2 = ... = ak = 1 ∼ I G = Zp1 × Zp2 × ... × Zpk

Automorphism Original problem (continued) groups Gerhardt Hinkle

I |G| = p1p2...pk and |Aut(G)| = (p1 − 1)(p2 − 1)...(pk − 1) I |Aut(G)| < |G|, so |Aut(G)| − |G| = 1 is impossible.

I If (p1 − 1)(p2 − 1)...(pk − 1) − p1p2...pk = −1, then k = 1.

I Therefore, d(G) = 1 is impossible, and d(G) = −1 if ∼ and only if G = Zp for some prime p.

Automorphism Original problem (continued) groups Gerhardt Hinkle

I What if two of the primes are the same? Introduction Automorphisms Original problem Generalizations Prime difference Prime square difference Possible differences Further research 2 2 I |Aut(Zp × Zp)| = (p − 1)(p − p) I Then, p divides |Aut(G)| and |G|, a contradiction. ∼ I Therefore, G = Zp1 × Zp2 × ... × Zpk for distinct primes p1, p2, ..., pk .

I |G| = p1p2...pk and |Aut(G)| = (p1 − 1)(p2 − 1)...(pk − 1) I |Aut(G)| < |G|, so |Aut(G)| − |G| = 1 is impossible.

I If (p1 − 1)(p2 − 1)...(pk − 1) − p1p2...pk = −1, then k = 1.

I Therefore, d(G) = 1 is impossible, and d(G) = −1 if ∼ and only if G = Zp for some prime p.

Automorphism Original problem (continued) groups Gerhardt Hinkle

I What if two of the primes are the same? Introduction Automorphisms Original problem I Suppose G ≥ Zp × Zp. Generalizations Prime difference Prime square difference Possible differences Further research I Then, p divides |Aut(G)| and |G|, a contradiction. ∼ I Therefore, G = Zp1 × Zp2 × ... × Zpk for distinct primes p1, p2, ..., pk .

I |G| = p1p2...pk and |Aut(G)| = (p1 − 1)(p2 − 1)...(pk − 1) I |Aut(G)| < |G|, so |Aut(G)| − |G| = 1 is impossible.

I If (p1 − 1)(p2 − 1)...(pk − 1) − p1p2...pk = −1, then k = 1.

I Therefore, d(G) = 1 is impossible, and d(G) = −1 if ∼ and only if G = Zp for some prime p.

Automorphism Original problem (continued) groups Gerhardt Hinkle

I What if two of the primes are the same? Introduction Automorphisms Original problem I Suppose G ≥ Zp × Zp. Generalizations 2 2 I |Aut( × )| = (p − 1)(p − p) Prime difference Zp Zp Prime square difference Possible differences Further research ∼ I Therefore, G = Zp1 × Zp2 × ... × Zpk for distinct primes p1, p2, ..., pk .

I |G| = p1p2...pk and |Aut(G)| = (p1 − 1)(p2 − 1)...(pk − 1) I |Aut(G)| < |G|, so |Aut(G)| − |G| = 1 is impossible.

I If (p1 − 1)(p2 − 1)...(pk − 1) − p1p2...pk = −1, then k = 1.

I Therefore, d(G) = 1 is impossible, and d(G) = −1 if ∼ and only if G = Zp for some prime p.

Automorphism Original problem (continued) groups Gerhardt Hinkle

I What if two of the primes are the same? Introduction Automorphisms Original problem I Suppose G ≥ Zp × Zp. Generalizations 2 2 I |Aut( × )| = (p − 1)(p − p) Prime difference Zp Zp Prime square difference I Then, p divides |Aut(G)| and |G|, a contradiction. Possible differences Further research I |G| = p1p2...pk and |Aut(G)| = (p1 − 1)(p2 − 1)...(pk − 1) I |Aut(G)| < |G|, so |Aut(G)| − |G| = 1 is impossible.

I If (p1 − 1)(p2 − 1)...(pk − 1) − p1p2...pk = −1, then k = 1.

I Therefore, d(G) = 1 is impossible, and d(G) = −1 if ∼ and only if G = Zp for some prime p.

Automorphism Original problem (continued) groups Gerhardt Hinkle

I What if two of the primes are the same? Introduction Automorphisms Original problem I Suppose G ≥ Zp × Zp. Generalizations 2 2 I |Aut( × )| = (p − 1)(p − p) Prime difference Zp Zp Prime square difference I Then, p divides |Aut(G)| and |G|, a contradiction. Possible differences ∼ Further research I Therefore, G = Zp1 × Zp2 × ... × Zpk for distinct primes p1, p2, ..., pk . I |Aut(G)| < |G|, so |Aut(G)| − |G| = 1 is impossible.

I If (p1 − 1)(p2 − 1)...(pk − 1) − p1p2...pk = −1, then k = 1.

I Therefore, d(G) = 1 is impossible, and d(G) = −1 if ∼ and only if G = Zp for some prime p.

Automorphism Original problem (continued) groups Gerhardt Hinkle

I What if two of the primes are the same? Introduction Automorphisms Original problem I Suppose G ≥ Zp × Zp. Generalizations 2 2 I |Aut( × )| = (p − 1)(p − p) Prime difference Zp Zp Prime square difference I Then, p divides |Aut(G)| and |G|, a contradiction. Possible differences ∼ Further research I Therefore, G = Zp1 × Zp2 × ... × Zpk for distinct primes p1, p2, ..., pk .

I |G| = p1p2...pk and |Aut(G)| = (p1 − 1)(p2 − 1)...(pk − 1) I If (p1 − 1)(p2 − 1)...(pk − 1) − p1p2...pk = −1, then k = 1.

I Therefore, d(G) = 1 is impossible, and d(G) = −1 if ∼ and only if G = Zp for some prime p.

Automorphism Original problem (continued) groups Gerhardt Hinkle

I What if two of the primes are the same? Introduction Automorphisms Original problem I Suppose G ≥ Zp × Zp. Generalizations 2 2 I |Aut( × )| = (p − 1)(p − p) Prime difference Zp Zp Prime square difference I Then, p divides |Aut(G)| and |G|, a contradiction. Possible differences ∼ Further research I Therefore, G = Zp1 × Zp2 × ... × Zpk for distinct primes p1, p2, ..., pk .

I |G| = p1p2...pk and |Aut(G)| = (p1 − 1)(p2 − 1)...(pk − 1) I |Aut(G)| < |G|, so |Aut(G)| − |G| = 1 is impossible. I Therefore, d(G) = 1 is impossible, and d(G) = −1 if ∼ and only if G = Zp for some prime p.

Automorphism Original problem (continued) groups Gerhardt Hinkle

I What if two of the primes are the same? Introduction Automorphisms Original problem I Suppose G ≥ Zp × Zp. Generalizations 2 2 I |Aut( × )| = (p − 1)(p − p) Prime difference Zp Zp Prime square difference I Then, p divides |Aut(G)| and |G|, a contradiction. Possible differences ∼ Further research I Therefore, G = Zp1 × Zp2 × ... × Zpk for distinct primes p1, p2, ..., pk .

I |G| = p1p2...pk and |Aut(G)| = (p1 − 1)(p2 − 1)...(pk − 1) I |Aut(G)| < |G|, so |Aut(G)| − |G| = 1 is impossible.

I If (p1 − 1)(p2 − 1)...(pk − 1) − p1p2...pk = −1, then k = 1. Automorphism Original problem (continued) groups Gerhardt Hinkle

I What if two of the primes are the same? Introduction Automorphisms Original problem I Suppose G ≥ Zp × Zp. Generalizations 2 2 I |Aut( × )| = (p − 1)(p − p) Prime difference Zp Zp Prime square difference I Then, p divides |Aut(G)| and |G|, a contradiction. Possible differences ∼ Further research I Therefore, G = Zp1 × Zp2 × ... × Zpk for distinct primes p1, p2, ..., pk .

I |G| = p1p2...pk and |Aut(G)| = (p1 − 1)(p2 − 1)...(pk − 1) I |Aut(G)| < |G|, so |Aut(G)| − |G| = 1 is impossible.

I If (p1 − 1)(p2 − 1)...(pk − 1) − p1p2...pk = −1, then k = 1.

I Therefore, d(G) = 1 is impossible, and d(G) = −1 if ∼ and only if G = Zp for some prime p. 2 I What groups give d(G) = ±p ?

I What values of d(G) are possible?

Automorphism Generalizations groups Gerhardt Hinkle

Introduction Automorphisms Original problem Generalizations Prime difference Prime square difference Possible differences I What groups give d(G) = ±p? Further research I What values of d(G) are possible?

Automorphism Generalizations groups Gerhardt Hinkle

Introduction Automorphisms Original problem Generalizations Prime difference Prime square difference Possible differences I What groups give d(G) = ±p? Further research 2 I What groups give d(G) = ±p ? Automorphism Generalizations groups Gerhardt Hinkle

I What values of d(G) are possible? ∼ I G = Zq1 × Zq2 × ... × Zqk ∼ I G = Zp × Zq1 × Zq2 × ... × Zqk ∼ 2 I G = Zp × Zq1 × Zq2 × ... × Zqk ∼ I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk

I |Inn(G)| = 1, p I The only group of order p is Zp, but it is impossible for G/Z(G) to be a nontrivial cyclic group.

I Therefore, |Inn(G)| = 1, so G is abelian.

I By a similar argument to the d(G) = ±1 case, all prime factors must be distinct and have exponent 1, except that there could be either two ps or one p2 (but not both).

I Possible cases: (q1, q2, ...qk distinct primes not equal to p)

Automorphism Prime diﬀerence groups Gerhardt Hinkle

I d(G) = |Aut(G)| − |G| = ±p Introduction Automorphisms Original problem Generalizations Prime difference Prime square difference Possible differences Further research ∼ I G = Zq1 × Zq2 × ... × Zqk ∼ I G = Zp × Zq1 × Zq2 × ... × Zqk ∼ 2 I G = Zp × Zq1 × Zq2 × ... × Zqk ∼ I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk

I The only group of order p is Zp, but it is impossible for G/Z(G) to be a nontrivial cyclic group.

I Therefore, |Inn(G)| = 1, so G is abelian.

I By a similar argument to the d(G) = ±1 case, all prime factors must be distinct and have exponent 1, except that there could be either two ps or one p2 (but not both).

I Possible cases: (q1, q2, ...qk distinct primes not equal to p)

Automorphism Prime diﬀerence groups Gerhardt Hinkle

I d(G) = |Aut(G)| − |G| = ±p Introduction Automorphisms I |Inn(G)| = 1, p Original problem Generalizations Prime difference Prime square difference Possible differences Further research ∼ I G = Zq1 × Zq2 × ... × Zqk ∼ I G = Zp × Zq1 × Zq2 × ... × Zqk ∼ 2 I G = Zp × Zq1 × Zq2 × ... × Zqk ∼ I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk

I Therefore, |Inn(G)| = 1, so G is abelian.

I By a similar argument to the d(G) = ±1 case, all prime factors must be distinct and have exponent 1, except that there could be either two ps or one p2 (but not both).

I Possible cases: (q1, q2, ...qk distinct primes not equal to p)

Automorphism Prime diﬀerence groups Gerhardt Hinkle

I d(G) = |Aut(G)| − |G| = ±p Introduction Automorphisms I |Inn(G)| = 1, p Original problem Generalizations I The only group of order p is Zp, but it is impossible for Prime difference Prime square G/Z(G) to be a nontrivial cyclic group. difference Possible differences Further research ∼ I G = Zq1 × Zq2 × ... × Zqk ∼ I G = Zp × Zq1 × Zq2 × ... × Zqk ∼ 2 I G = Zp × Zq1 × Zq2 × ... × Zqk ∼ I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk

I By a similar argument to the d(G) = ±1 case, all prime factors must be distinct and have exponent 1, except that there could be either two ps or one p2 (but not both).

I Possible cases: (q1, q2, ...qk distinct primes not equal to p)

Automorphism Prime diﬀerence groups Gerhardt Hinkle

I d(G) = |Aut(G)| − |G| = ±p Introduction Automorphisms I |Inn(G)| = 1, p Original problem Generalizations I The only group of order p is Zp, but it is impossible for Prime difference Prime square G/Z(G) to be a nontrivial cyclic group. difference Possible differences I Therefore, |Inn(G)| = 1, so G is abelian. Further research ∼ I G = Zq1 × Zq2 × ... × Zqk ∼ I G = Zp × Zq1 × Zq2 × ... × Zqk ∼ 2 I G = Zp × Zq1 × Zq2 × ... × Zqk ∼ I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk

I Possible cases: (q1, q2, ...qk distinct primes not equal to p)

Automorphism Prime diﬀerence groups Gerhardt Hinkle

I By a similar argument to the d(G) = ±1 case, all prime factors must be distinct and have exponent 1, except that there could be either two ps or one p2 (but not both). ∼ I G = Zq1 × Zq2 × ... × Zqk ∼ I G = Zp × Zq1 × Zq2 × ... × Zqk ∼ 2 I G = Zp × Zq1 × Zq2 × ... × Zqk ∼ I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk

Automorphism Prime diﬀerence groups Gerhardt Hinkle

I By a similar argument to the d(G) = ±1 case, all prime factors must be distinct and have exponent 1, except that there could be either two ps or one p2 (but not both).

I Possible cases: (q1, q2, ...qk distinct primes not equal to p) ∼ I G = Zp × Zq1 × Zq2 × ... × Zqk ∼ 2 I G = Zp × Zq1 × Zq2 × ... × Zqk ∼ I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk

Automorphism Prime diﬀerence groups Gerhardt Hinkle

I By a similar argument to the d(G) = ±1 case, all prime factors must be distinct and have exponent 1, except that there could be either two ps or one p2 (but not both).

I Possible cases: (q1, q2, ...qk distinct primes not equal to p) ∼ I G = Zq1 × Zq2 × ... × Zqk ∼ 2 I G = Zp × Zq1 × Zq2 × ... × Zqk ∼ I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk

Automorphism Prime diﬀerence groups Gerhardt Hinkle

I By a similar argument to the d(G) = ±1 case, all prime factors must be distinct and have exponent 1, except that there could be either two ps or one p2 (but not both).

I Possible cases: (q1, q2, ...qk distinct primes not equal to p) ∼ I G = Zq1 × Zq2 × ... × Zqk ∼ I G = Zp × Zq1 × Zq2 × ... × Zqk ∼ I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk

Automorphism Prime diﬀerence groups Gerhardt Hinkle

I By a similar argument to the d(G) = ±1 case, all prime factors must be distinct and have exponent 1, except that there could be either two ps or one p2 (but not both).

I Possible cases: (q1, q2, ...qk distinct primes not equal to p) ∼ I G = Zq1 × Zq2 × ... × Zqk ∼ I G = Zp × Zq1 × Zq2 × ... × Zqk ∼ 2 I G = Zp × Zq1 × Zq2 × ... × Zqk Automorphism Prime diﬀerence groups Gerhardt Hinkle

I By a similar argument to the d(G) = ±1 case, all prime factors must be distinct and have exponent 1, except that there could be either two ps or one p2 (but not both).

I Possible cases: (q1, q2, ...qk distinct primes not equal to p) ∼ I G = Zq1 × Zq2 × ... × Zqk ∼ I G = Zp × Zq1 × Zq2 × ... × Zqk ∼ 2 I G = Zp × Zq1 × Zq2 × ... × Zqk ∼ I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk I k = 2: p ≥ 7 I k = 3: p ≥ 57 I k = 4: p ≥ 675

I (q1 − 1)(q2 − 1)...(qk − 1) − q1q2...qk = ±p

I (q1 − 1)(q2 − 1)...(qk − 1) − q1q2...qk = −p I There is no general form for the solutions, although they appear to exist for all p.

I k = 2: q1 + q2 = p + 1 I Increasing any qi increases the magnitude of the diﬀerence, so the lower bound for what values of p can be obtained for a given k is 3 · 5 · 7 · ... · pk+1 − 2 · 4 · 6 · ... · (pk+1 − 1). This can be reversed to get an upper bound on k for a given p.

Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

∼ Introduction I G = Zq1 × Zq2 × ... × Zqk Automorphisms Original problem Generalizations Prime difference Prime square difference Possible differences Further research I k = 2: p ≥ 7 I k = 3: p ≥ 57 I k = 4: p ≥ 675

I (q1 − 1)(q2 − 1)...(qk − 1) − q1q2...qk = −p I There is no general form for the solutions, although they appear to exist for all p.

Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

∼ Introduction I G = Zq1 × Zq2 × ... × Zqk Automorphisms Original problem I (q1 − 1)(q2 − 1)...(qk − 1) − q1q2...qk = ±p Generalizations Prime difference Prime square difference Possible differences Further research I k = 2: p ≥ 7 I k = 3: p ≥ 57 I k = 4: p ≥ 675

I There is no general form for the solutions, although they appear to exist for all p.

Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

∼ Introduction I G = Zq1 × Zq2 × ... × Zqk Automorphisms Original problem I (q1 − 1)(q2 − 1)...(qk − 1) − q1q2...qk = ±p Generalizations Prime difference I (q1 − 1)(q2 − 1)...(qk − 1) − q1q2...qk = −p Prime square difference Possible differences Further research I k = 2: p ≥ 7 I k = 3: p ≥ 57 I k = 4: p ≥ 675

Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

∼ Introduction I G = Zq1 × Zq2 × ... × Zqk Automorphisms Original problem I (q1 − 1)(q2 − 1)...(qk − 1) − q1q2...qk = ±p Generalizations Prime difference I (q1 − 1)(q2 − 1)...(qk − 1) − q1q2...qk = −p Prime square difference I There is no general form for the solutions, although Possible differences they appear to exist for all p. Further research I k = 2: p ≥ 7 I k = 3: p ≥ 57 I k = 4: p ≥ 675

I Increasing any qi increases the magnitude of the diﬀerence, so the lower bound for what values of p can be obtained for a given k is 3 · 5 · 7 · ... · pk+1 − 2 · 4 · 6 · ... · (pk+1 − 1). This can be reversed to get an upper bound on k for a given p.

Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

∼ Introduction I G = Zq1 × Zq2 × ... × Zqk Automorphisms Original problem I (q1 − 1)(q2 − 1)...(qk − 1) − q1q2...qk = ±p Generalizations Prime difference I (q1 − 1)(q2 − 1)...(qk − 1) − q1q2...qk = −p Prime square difference I There is no general form for the solutions, although Possible differences they appear to exist for all p. Further research

I k = 2: q1 + q2 = p + 1 I k = 2: p ≥ 7 I k = 3: p ≥ 57 I k = 4: p ≥ 675

Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

∼ Introduction I G = Zq1 × Zq2 × ... × Zqk Automorphisms Original problem I (q1 − 1)(q2 − 1)...(qk − 1) − q1q2...qk = ±p Generalizations Prime difference I (q1 − 1)(q2 − 1)...(qk − 1) − q1q2...qk = −p Prime square difference I There is no general form for the solutions, although Possible differences they appear to exist for all p. Further research

Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

∼ Introduction I G = Zq1 × Zq2 × ... × Zqk Automorphisms Original problem I (q1 − 1)(q2 − 1)...(qk − 1) − q1q2...qk = ±p Generalizations Prime difference I (q1 − 1)(q2 − 1)...(qk − 1) − q1q2...qk = −p Prime square difference I There is no general form for the solutions, although Possible differences they appear to exist for all p. Further research

I k = 2: p ≥ 7 I k = 4: p ≥ 675

Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

∼ Introduction I G = Zq1 × Zq2 × ... × Zqk Automorphisms Original problem I (q1 − 1)(q2 − 1)...(qk − 1) − q1q2...qk = ±p Generalizations Prime difference I (q1 − 1)(q2 − 1)...(qk − 1) − q1q2...qk = −p Prime square difference I There is no general form for the solutions, although Possible differences they appear to exist for all p. Further research

I k = 2: p ≥ 7 I k = 3: p ≥ 57 Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

∼ Introduction I G = Zq1 × Zq2 × ... × Zqk Automorphisms Original problem I (q1 − 1)(q2 − 1)...(qk − 1) − q1q2...qk = ±p Generalizations Prime difference I (q1 − 1)(q2 − 1)...(qk − 1) − q1q2...qk = −p Prime square difference I There is no general form for the solutions, although Possible differences they appear to exist for all p. Further research

I k = 2: p ≥ 7 I k = 3: p ≥ 57 I k = 4: p ≥ 675 I (p − 1)(q1 − 1)(q2 − 1)...(qk − 1) − pq1q2...qk = ±p

I (p − 1)(q1 − 1)(q2 − 1)...(qk − 1) − pq1q2...qk = −p

I p(q1q2...qk − (q1 − 1)(q2 − 1)...(qk − 1) − 1) + (q1 − 1)(q2 − 1)...(qk − 1) = 0 I Both terms on the left side are positive, so there is no solution.

Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

Introduction Automorphisms Original problem Generalizations Prime difference I G ∼ × × × ... × Prime square = Zp Zq1 Zq2 Zqk difference Possible differences Further research I (p − 1)(q1 − 1)(q2 − 1)...(qk − 1) − pq1q2...qk = −p

I p(q1q2...qk − (q1 − 1)(q2 − 1)...(qk − 1) − 1) + (q1 − 1)(q2 − 1)...(qk − 1) = 0 I Both terms on the left side are positive, so there is no solution.

Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

Introduction Automorphisms Original problem Generalizations Prime difference I G ∼ × × × ... × Prime square = Zp Zq1 Zq2 Zqk difference Possible differences I (p − 1)(q1 − 1)(q2 − 1)...(qk − 1) − pq1q2...qk = ±p Further research I p(q1q2...qk − (q1 − 1)(q2 − 1)...(qk − 1) − 1) + (q1 − 1)(q2 − 1)...(qk − 1) = 0 I Both terms on the left side are positive, so there is no solution.

Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

Introduction Automorphisms Original problem Generalizations Prime difference I G ∼ × × × ... × Prime square = Zp Zq1 Zq2 Zqk difference Possible differences I (p − 1)(q1 − 1)(q2 − 1)...(qk − 1) − pq1q2...qk = ±p Further research I (p − 1)(q1 − 1)(q2 − 1)...(qk − 1) − pq1q2...qk = −p I Both terms on the left side are positive, so there is no solution.

Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

Introduction Automorphisms Original problem Generalizations Prime difference I G ∼ × × × ... × Prime square = Zp Zq1 Zq2 Zqk difference Possible differences I (p − 1)(q1 − 1)(q2 − 1)...(qk − 1) − pq1q2...qk = ±p Further research I (p − 1)(q1 − 1)(q2 − 1)...(qk − 1) − pq1q2...qk = −p

I p(q1q2...qk − (q1 − 1)(q2 − 1)...(qk − 1) − 1) + (q1 − 1)(q2 − 1)...(qk − 1) = 0 Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

Introduction Automorphisms Original problem Generalizations Prime difference I G ∼ × × × ... × Prime square = Zp Zq1 Zq2 Zqk difference Possible differences I (p − 1)(q1 − 1)(q2 − 1)...(qk − 1) − pq1q2...qk = ±p Further research I (p − 1)(q1 − 1)(q2 − 1)...(qk − 1) − pq1q2...qk = −p

I p(q1q2...qk − (q1 − 1)(q2 − 1)...(qk − 1) − 1) + (q1 − 1)(q2 − 1)...(qk − 1) = 0 I Both terms on the left side are positive, so there is no solution. 2 2 I (p − p)(q1 − 1)(q2 − 1)...(qk − 1) − p q1q2...qk = ±p 2 2 I (p − p)(q1 − 1)(q2 − 1)...(qk − 1) − p q1q2...qk = −p

I (p − 1)(q1 − 1)(q − 2 − 1)...(qk − 1) − pq1q2...qk = −1 I Only possible if k = 0 ∼ I G = Zp2

Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

Introduction Automorphisms Original problem Generalizations Prime difference ∼ Prime square I G = Zp2 × Zq1 × Zq2 × ... × Zqk difference Possible differences Further research 2 2 I (p − p)(q1 − 1)(q2 − 1)...(qk − 1) − p q1q2...qk = −p

I (p − 1)(q1 − 1)(q − 2 − 1)...(qk − 1) − pq1q2...qk = −1 I Only possible if k = 0 ∼ I G = Zp2

Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

Introduction Automorphisms Original problem Generalizations Prime difference ∼ Prime square I G = Zp2 × Zq1 × Zq2 × ... × Zqk difference 2 2 Possible differences I (p − p)(q1 − 1)(q2 − 1)...(qk − 1) − p q1q2...qk = ±p Further research I (p − 1)(q1 − 1)(q − 2 − 1)...(qk − 1) − pq1q2...qk = −1 I Only possible if k = 0 ∼ I G = Zp2

Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

Introduction Automorphisms Original problem Generalizations Prime difference ∼ Prime square I G = Zp2 × Zq1 × Zq2 × ... × Zqk difference 2 2 Possible differences I (p − p)(q1 − 1)(q2 − 1)...(qk − 1) − p q1q2...qk = ±p Further research 2 2 I (p − p)(q1 − 1)(q2 − 1)...(qk − 1) − p q1q2...qk = −p I Only possible if k = 0 ∼ I G = Zp2

Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

Introduction Automorphisms Original problem Generalizations Prime difference ∼ Prime square I G = Zp2 × Zq1 × Zq2 × ... × Zqk difference 2 2 Possible differences I (p − p)(q1 − 1)(q2 − 1)...(qk − 1) − p q1q2...qk = ±p Further research 2 2 I (p − p)(q1 − 1)(q2 − 1)...(qk − 1) − p q1q2...qk = −p

I (p − 1)(q1 − 1)(q − 2 − 1)...(qk − 1) − pq1q2...qk = −1 ∼ I G = Zp2

Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

Introduction Automorphisms Original problem Generalizations Prime difference ∼ Prime square I G = Zp2 × Zq1 × Zq2 × ... × Zqk difference 2 2 Possible differences I (p − p)(q1 − 1)(q2 − 1)...(qk − 1) − p q1q2...qk = ±p Further research 2 2 I (p − p)(q1 − 1)(q2 − 1)...(qk − 1) − p q1q2...qk = −p

I (p − 1)(q1 − 1)(q − 2 − 1)...(qk − 1) − pq1q2...qk = −1 I Only possible if k = 0 Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

Introduction Automorphisms Original problem Generalizations Prime difference ∼ Prime square I G = Zp2 × Zq1 × Zq2 × ... × Zqk difference 2 2 Possible differences I (p − p)(q1 − 1)(q2 − 1)...(qk − 1) − p q1q2...qk = ±p Further research 2 2 I (p − p)(q1 − 1)(q2 − 1)...(qk − 1) − p q1q2...qk = −p

I (p − 1)(q1 − 1)(q − 2 − 1)...(qk − 1) − pq1q2...qk = −1 I Only possible if k = 0 ∼ I G = Zp2 2 2 2 I (p −1)(p −p)(q1 −1)(q2 −1)...(qk −1)−p q1q2...qk = ±p 2 I (p −1)(p−1)(q1−1)(q2−1)...(qk −1)−pq1q2...qk = ±1 I p = 2 has no solution, so p ≥ 3.

Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

Introduction Automorphisms ∼ Original problem I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk Generalizations Prime difference Prime square difference Possible differences Further research 2 I (p −1)(p−1)(q1−1)(q2−1)...(qk −1)−pq1q2...qk = ±1 I p = 2 has no solution, so p ≥ 3.

Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

Introduction Automorphisms ∼ Original problem I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk 2 2 2 Generalizations I (p −1)(p −p)(q1 −1)(q2 −1)...(qk −1)−p q1q2...qk = Prime difference Prime square ±p difference Possible differences 2 I (p −1)(p−1)(q1−1)(q2−1)...(qk −1)−pq1q2...qk = ±1 Further research I f (p) = 2 (p −1)(p −1)(q1 −1)(q2 −1)...(qk −1)−pq1q2...qk ∓1 0 I There are no solutions if f (3) > 0 and f (p) > 0 for all p ≥ 3. 0 I There are no solutions if f (3) > 0, f (3) > 0, and f 00(p) > 0 for all p ≥ 3.

Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

Introduction Automorphisms ∼ Original problem I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk 2 2 2 Generalizations I (p −1)(p −p)(q1 −1)(q2 −1)...(qk −1)−p q1q2...qk = Prime difference Prime square ±p difference Possible differences 2 I (p −1)(p−1)(q1−1)(q2−1)...(qk −1)−pq1q2...qk = ±1 Further research I p = 2 has no solution, so p ≥ 3. 0 I There are no solutions if f (3) > 0 and f (p) > 0 for all p ≥ 3. 0 I There are no solutions if f (3) > 0, f (3) > 0, and f 00(p) > 0 for all p ≥ 3.

Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

I f (p) = 2 (p −1)(p −1)(q1 −1)(q2 −1)...(qk −1)−pq1q2...qk ∓1 0 I There are no solutions if f (3) > 0, f (3) > 0, and f 00(p) > 0 for all p ≥ 3.

Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

I f (p) = 2 (p −1)(p −1)(q1 −1)(q2 −1)...(qk −1)−pq1q2...qk ∓1 0 I There are no solutions if f (3) > 0 and f (p) > 0 for all p ≥ 3. Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

I f (p) = 2 (p −1)(p −1)(q1 −1)(q2 −1)...(qk −1)−pq1q2...qk ∓1 0 I There are no solutions if f (3) > 0 and f (p) > 0 for all p ≥ 3. 0 I There are no solutions if f (3) > 0, f (3) > 0, and f 00(p) > 0 for all p ≥ 3. 0 I f (3) = 20(q1 − 1)(q2 − 1)...(qk − 1) − q1q2...qk 00 I f (p) = (6p − 2)(q1 − 1)(q2 − 1)...(qk − 1) 00 I f (p) > 0 for all p ≥ 3 always holds.

I No solutions if 16(q1 − 1)(q2 − 1)...(qk − 1) − 3q1q2...qk > 0, unless 16(q1 − 1)(q2 − 1)...(qk − 1) − 3q1q2...qk = 1, in which case d(G) = p and p = 3

I If 16(q1 − 1)(q2 − 1)...(qk − 1) − 3q1q2...qk > 0, then 20(q1 − 1)(q2 − 1)...(qk − 1) − q1q2...qk > 0.

Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

Introduction Automorphisms Original problem I f (3) = 16(q1 − 1)(q2 − 1)...(qk − 1) − 3q1q2...qk ∓ 1 Generalizations Prime difference Prime square difference Possible differences Further research 00 I f (p) = (6p − 2)(q1 − 1)(q2 − 1)...(qk − 1) 00 I f (p) > 0 for all p ≥ 3 always holds.

I No solutions if 16(q1 − 1)(q2 − 1)...(qk − 1) − 3q1q2...qk > 0, unless 16(q1 − 1)(q2 − 1)...(qk − 1) − 3q1q2...qk = 1, in which case d(G) = p and p = 3

I If 16(q1 − 1)(q2 − 1)...(qk − 1) − 3q1q2...qk > 0, then 20(q1 − 1)(q2 − 1)...(qk − 1) − q1q2...qk > 0.

Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

I No solutions if 16(q1 − 1)(q2 − 1)...(qk − 1) − 3q1q2...qk > 0, unless 16(q1 − 1)(q2 − 1)...(qk − 1) − 3q1q2...qk = 1, in which case d(G) = p and p = 3

I If 16(q1 − 1)(q2 − 1)...(qk − 1) − 3q1q2...qk > 0, then 20(q1 − 1)(q2 − 1)...(qk − 1) − q1q2...qk > 0.

Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

Introduction Automorphisms Original problem I f (3) = 16(q1 − 1)(q2 − 1)...(qk − 1) − 3q1q2...qk ∓ 1 Generalizations 0 Prime difference I f (3) = 20(q1 − 1)(q2 − 1)...(qk − 1) − q1q2...qk Prime square difference 00 Possible differences I f (p) = (6p − 2)(q1 − 1)(q2 − 1)...(qk − 1) Further research I No solutions if 16(q1 − 1)(q2 − 1)...(qk − 1) − 3q1q2...qk > 0, unless 16(q1 − 1)(q2 − 1)...(qk − 1) − 3q1q2...qk = 1, in which case d(G) = p and p = 3

I If 16(q1 − 1)(q2 − 1)...(qk − 1) − 3q1q2...qk > 0, then 20(q1 − 1)(q2 − 1)...(qk − 1) − q1q2...qk > 0.

Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

Introduction Automorphisms Original problem I f (3) = 16(q1 − 1)(q2 − 1)...(qk − 1) − 3q1q2...qk ∓ 1 Generalizations 0 Prime difference I f (3) = 20(q1 − 1)(q2 − 1)...(qk − 1) − q1q2...qk Prime square difference 00 Possible differences I f (p) = (6p − 2)(q1 − 1)(q2 − 1)...(qk − 1) Further research 00 I f (p) > 0 for all p ≥ 3 always holds. I If 16(q1 − 1)(q2 − 1)...(qk − 1) − 3q1q2...qk > 0, then 20(q1 − 1)(q2 − 1)...(qk − 1) − q1q2...qk > 0.

Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

I No solutions if 16(q1 − 1)(q2 − 1)...(qk − 1) − 3q1q2...qk > 0, unless 16(q1 − 1)(q2 − 1)...(qk − 1) − 3q1q2...qk = 1, in which case d(G) = p and p = 3 Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

I No solutions if 16(q1 − 1)(q2 − 1)...(qk − 1) − 3q1q2...qk > 0, unless 16(q1 − 1)(q2 − 1)...(qk − 1) − 3q1q2...qk = 1, in which case d(G) = p and p = 3

I If 16(q1 − 1)(q2 − 1)...(qk − 1) − 3q1q2...qk > 0, then 20(q1 − 1)(q2 − 1)...(qk − 1) − q1q2...qk > 0. 1 1 1 3 I 1 − 1 − ... 1 − ≤ q1 q2 qk 16 I If q1 = 2, 3, then there are no solutions.

I Minimum value when q1 = 5, q2 = 7, ..., qk = pk+2, where pk+2 is the (k + 2)th prime 1 1 1 3 I 1 − 1 − ... 1 − ≤ 5 7 pk+2 16 I k ≥ 994

I Other possibility: k = 993, p = 3, ∼ G = Z3 × Z3 × Z5 × Z7 × ... × Zp995 , and d(G) = 3 (can be easily conﬁrmed to be false)

I Therefore, a solution to d(G) = ±p exists if and only if k ≥ 994.

I Minimum value when q1 = 5, q2 = 7, ..., qk = pk+2, where pk+2 is the (k + 2)th prime 1 1 1 3 I 1 − 1 − ... 1 − ≤ 5 7 pk+2 16 I k ≥ 994

I Other possibility: k = 993, p = 3, ∼ G = Z3 × Z3 × Z5 × Z7 × ... × Zp995 , and d(G) = 3 (can be easily conﬁrmed to be false)

I Therefore, a solution to d(G) = ±p exists if and only if k ≥ 994.

Automorphism Prime difference (continued) groups Gerhardt Hinkle I There are only solutions if Introduction 16(q1 − 1)(q2 − 1)...(qk − 1) − 3q1q2...qk ≤ 0 Automorphisms Original problem (excluding the one previously-mentioned exception). Generalizations 1 1 1 3 Prime difference I 1 − 1 − ... 1 − ≤ Prime square q1 q2 qk 16 difference Possible differences Further research I Minimum value when q1 = 5, q2 = 7, ..., qk = pk+2, where pk+2 is the (k + 2)th prime 1 1 1 3 I 1 − 1 − ... 1 − ≤ 5 7 pk+2 16 I k ≥ 994

I Other possibility: k = 993, p = 3, ∼ G = Z3 × Z3 × Z5 × Z7 × ... × Zp995 , and d(G) = 3 (can be easily conﬁrmed to be false)

I Therefore, a solution to d(G) = ±p exists if and only if k ≥ 994.

Automorphism Prime difference (continued) groups Gerhardt Hinkle I There are only solutions if Introduction 16(q1 − 1)(q2 − 1)...(qk − 1) − 3q1q2...qk ≤ 0 Automorphisms Original problem (excluding the one previously-mentioned exception). Generalizations 1 1 1 3 Prime difference I 1 − 1 − ... 1 − ≤ Prime square q1 q2 qk 16 difference Possible differences I If q1 = 2, 3, then there are no solutions. Further research 1 1 1 3 I 1 − 1 − ... 1 − ≤ 5 7 pk+2 16 I k ≥ 994

I Other possibility: k = 993, p = 3, ∼ G = Z3 × Z3 × Z5 × Z7 × ... × Zp995 , and d(G) = 3 (can be easily conﬁrmed to be false)

I Therefore, a solution to d(G) = ±p exists if and only if k ≥ 994.

Automorphism Prime difference (continued) groups Gerhardt Hinkle I There are only solutions if Introduction 16(q1 − 1)(q2 − 1)...(qk − 1) − 3q1q2...qk ≤ 0 Automorphisms Original problem (excluding the one previously-mentioned exception). Generalizations 1 1 1 3 Prime difference I 1 − 1 − ... 1 − ≤ Prime square q1 q2 qk 16 difference Possible differences I If q1 = 2, 3, then there are no solutions. Further research I Minimum value when q1 = 5, q2 = 7, ..., qk = pk+2, where pk+2 is the (k + 2)th prime I k ≥ 994

I Other possibility: k = 993, p = 3, ∼ G = Z3 × Z3 × Z5 × Z7 × ... × Zp995 , and d(G) = 3 (can be easily conﬁrmed to be false)

I Therefore, a solution to d(G) = ±p exists if and only if k ≥ 994.

Automorphism Prime difference (continued) groups Gerhardt Hinkle I There are only solutions if Introduction 16(q1 − 1)(q2 − 1)...(qk − 1) − 3q1q2...qk ≤ 0 Automorphisms Original problem (excluding the one previously-mentioned exception). Generalizations 1 1 1 3 Prime difference I 1 − 1 − ... 1 − ≤ Prime square q1 q2 qk 16 difference Possible differences I If q1 = 2, 3, then there are no solutions. Further research I Minimum value when q1 = 5, q2 = 7, ..., qk = pk+2, where pk+2 is the (k + 2)th prime 1 1 1 3 I 1 − 1 − ... 1 − ≤ 5 7 pk+2 16 I Other possibility: k = 993, p = 3, ∼ G = Z3 × Z3 × Z5 × Z7 × ... × Zp995 , and d(G) = 3 (can be easily confirmed to be false)

I Therefore, a solution to d(G) = ±p exists if and only if k ≥ 994.

Automorphism Prime difference (continued) groups Gerhardt Hinkle I There are only solutions if Introduction 16(q1 − 1)(q2 − 1)...(qk − 1) − 3q1q2...qk ≤ 0 Automorphisms Original problem (excluding the one previously-mentioned exception). Generalizations 1 1 1 3 Prime difference I 1 − 1 − ... 1 − ≤ Prime square q1 q2 qk 16 difference Possible differences I If q1 = 2, 3, then there are no solutions. Further research I Minimum value when q1 = 5, q2 = 7, ..., qk = pk+2, where pk+2 is the (k + 2)th prime 1 1 1 3 I 1 − 1 − ... 1 − ≤ 5 7 pk+2 16 I k ≥ 994 I Therefore, a solution to d(G) = ±p exists if and only if k ≥ 994.

Automorphism Prime difference (continued) groups Gerhardt Hinkle I There are only solutions if Introduction 16(q1 − 1)(q2 − 1)...(qk − 1) − 3q1q2...qk ≤ 0 Automorphisms Original problem (excluding the one previously-mentioned exception). Generalizations 1 1 1 3 Prime difference I 1 − 1 − ... 1 − ≤ Prime square q1 q2 qk 16 difference Possible differences I If q1 = 2, 3, then there are no solutions. Further research I Minimum value when q1 = 5, q2 = 7, ..., qk = pk+2, where pk+2 is the (k + 2)th prime 1 1 1 3 I 1 − 1 − ... 1 − ≤ 5 7 pk+2 16 I k ≥ 994

I Other possibility: k = 993, p = 3, ∼ G = Z3 × Z3 × Z5 × Z7 × ... × Zp995 , and d(G) = 3 (can be easily confirmed to be false) Automorphism Prime difference (continued) groups Gerhardt Hinkle I There are only solutions if Introduction 16(q1 − 1)(q2 − 1)...(qk − 1) − 3q1q2...qk ≤ 0 Automorphisms Original problem (excluding the one previously-mentioned exception). Generalizations 1 1 1 3 Prime difference I 1 − 1 − ... 1 − ≤ Prime square q1 q2 qk 16 difference Possible differences I If q1 = 2, 3, then there are no solutions. Further research I Minimum value when q1 = 5, q2 = 7, ..., qk = pk+2, where pk+2 is the (k + 2)th prime 1 1 1 3 I 1 − 1 − ... 1 − ≤ 5 7 pk+2 16 I k ≥ 994

I Other possibility: k = 993, p = 3, ∼ G = Z3 × Z3 × Z5 × Z7 × ... × Zp995 , and d(G) = 3 (can be easily conﬁrmed to be false)

I Therefore, a solution to d(G) = ±p exists if and only if k ≥ 994. 1 1 1 p0 I 1 − 3 1 − 5 ... 1 − p ≤ 2 , where the k+2 (p0 −1)(p0−1) product excludes the term containing p0 so that there are k terms in total.

I The product in the left side of the inequality goes to 0 as k goes to ∞, so there will always exist a value of k so that the inequality is satisﬁed.

I Let k(p0) be the lowest value of k satisfying the inequality for a given p0. Then, any group G for which d(G) = ±p0 must have k ≥ k(p0), except that it could be possible to have k = k(p0) − 1, {q1, q2, ..., qk } = {3, 5, 7, ..., pk+2}\{p0}, and d(G) = p0.

Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

I We can use a similar manner to find a lower bound on Introduction Automorphisms the values of k that give a difference of at least ±p0. Original problem Generalizations Prime difference Prime square difference Possible differences Further research I The product in the left side of the inequality goes to 0 as k goes to ∞, so there will always exist a value of k so that the inequality is satisfied.

Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

I We can use a similar manner to find a lower bound on Introduction Automorphisms the values of k that give a difference of at least ±p0. Original problem Generalizations 1 1 1 p0 I 1 − 3 1 − 5 ... 1 − p ≤ 2 , where the Prime difference k+2 (p0 −1)(p0−1) Prime square difference product excludes the term containing p0 so that there Possible differences are k terms in total. Further research I Let k(p0) be the lowest value of k satisfying the inequality for a given p0. Then, any group G for which d(G) = ±p0 must have k ≥ k(p0), except that it could be possible to have k = k(p0) − 1, {q1, q2, ..., qk } = {3, 5, 7, ..., pk+2}\{p0}, and d(G) = p0.

Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

I The product in the left side of the inequality goes to 0 as k goes to ∞, so there will always exist a value of k so that the inequality is satisﬁed. Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

I The product in the left side of the inequality goes to 0 as k goes to ∞, so there will always exist a value of k so that the inequality is satisﬁed.

I k(5) is too large to compute easily. (much greater than 20 million)

I The numbers are so large that the chance of getting ±1 is very remote, so I conjecture that there are no solutions.

Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

Introduction Automorphisms Original problem Generalizations Prime difference Prime square I k(p0) increases very rapidly. difference Possible differences Further research I k(5) is too large to compute easily. (much greater than 20 million)

I The numbers are so large that the chance of getting ±1 is very remote, so I conjecture that there are no solutions.

Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

Introduction Automorphisms Original problem Generalizations Prime difference Prime square I k(p0) increases very rapidly. difference k(3) = 994 Possible differences I Further research I The numbers are so large that the chance of getting ±1 is very remote, so I conjecture that there are no solutions.

Automorphism Prime diﬀerence (continued) groups Gerhardt Hinkle

Introduction Automorphisms Original problem Generalizations Prime difference Prime square I k(p0) increases very rapidly. difference k(3) = 994 Possible differences I Further research I k(5) is too large to compute easily. (much greater than 20 million) Automorphism Prime difference (continued) groups Gerhardt Hinkle

Introduction Automorphisms Original problem Generalizations Prime difference Prime square I k(p0) increases very rapidly. difference k(3) = 994 Possible differences I Further research I k(5) is too large to compute easily. (much greater than 20 million)

I The numbers are so large that the chance of getting ±1 is very remote, so I conjecture that there are no solutions. 2 I |Inn(G)| = 1, p, p I The only group of order p is Zp, and the only groups of 2 order p are Zp2 and Zp × Zp. ∼ I Inn(G) = {e}, Zp × Zp ∼ I Either G is abelian or G/Z(G) = Zp × Zp.

Automorphism Prime square diﬀerence groups Gerhardt Hinkle

Introduction Automorphisms Original problem Generalizations Prime difference 2 Prime square I d(G) = |Aut(G)| − |G| = ±p difference Possible differences Further research I The only group of order p is Zp, and the only groups of 2 order p are Zp2 and Zp × Zp. ∼ I Inn(G) = {e}, Zp × Zp ∼ I Either G is abelian or G/Z(G) = Zp × Zp.

Automorphism Prime square diﬀerence groups Gerhardt Hinkle

Introduction Automorphisms Original problem Generalizations Prime difference 2 Prime square I d(G) = |Aut(G)| − |G| = ±p difference Possible differences 2 I |Inn(G)| = 1, p, p Further research I The only group of order p is Zp, and the only groups of 2 order p are Zp2 and Zp × Zp. ∼ I Either G is abelian or G/Z(G) = Zp × Zp.

Automorphism Prime square diﬀerence groups Gerhardt Hinkle

I Must be calculated manually

I No solution

∼ 2 I G = Zp3 , d(G) = −p

I k = 1: p = 2, q1 = 5, d(G) = 4 I k = 2: p = 2, q1 = 5, q2 = 7, d(G) = 4 I ... (must be calculated manually)

∼ I G = Zq1 × Zq2 × ... × Zqk

∼ I G = Zp × Zq1 × Zq2 × ... × Zqk

∼ 2 I G = Zp × Zq1 × Zq2 × ... × Zqk

∼ 3 I G = Zp × Zq1 × Zq2 × ... × Zqk

∼ I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk

I Possibilities:

Automorphism Abelian groups Gerhardt Hinkle

2 Introduction I d(G) = ±p , G abelian Automorphisms Original problem Generalizations Prime difference Prime square difference Possible differences Further research I Calculated like in the d(G) = ±p case; can only give d(G) = −p2

I Must be calculated manually

I No solution

∼ 2 I G = Zp3 , d(G) = −p

I k = 1: p = 2, q1 = 5, d(G) = 4 I k = 2: p = 2, q1 = 5, q2 = 7, d(G) = 4 I ... (must be calculated manually)

∼ I G = Zq1 × Zq2 × ... × Zqk

∼ I G = Zp × Zq1 × Zq2 × ... × Zqk

∼ 2 I G = Zp × Zq1 × Zq2 × ... × Zqk

∼ 3 I G = Zp × Zq1 × Zq2 × ... × Zqk

∼ I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk

Automorphism Abelian groups Gerhardt Hinkle

2 Introduction I d(G) = ±p , G abelian Automorphisms Original problem Possibilities: I Generalizations Prime difference Prime square difference Possible differences Further research I Must be calculated manually

I No solution

∼ 2 I G = Zp3 , d(G) = −p

I k = 1: p = 2, q1 = 5, d(G) = 4 I k = 2: p = 2, q1 = 5, q2 = 7, d(G) = 4 I ... (must be calculated manually)

I Calculated like in the d(G) = ±p case; can only give d(G) = −p2 ∼ I G = Zp × Zq1 × Zq2 × ... × Zqk

∼ 2 I G = Zp × Zq1 × Zq2 × ... × Zqk

∼ 3 I G = Zp × Zq1 × Zq2 × ... × Zqk

∼ I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk

Automorphism Abelian groups Gerhardt Hinkle

2 Introduction I d(G) = ±p , G abelian Automorphisms Original problem Possibilities: I Generalizations ∼ Prime difference I G = q1 × q2 × ... × qk Z Z Z Prime square difference Possible differences Further research I Must be calculated manually

I No solution

∼ 2 I G = Zp3 , d(G) = −p

I k = 1: p = 2, q1 = 5, d(G) = 4 I k = 2: p = 2, q1 = 5, q2 = 7, d(G) = 4 I ... (must be calculated manually)

∼ I G = Zp × Zq1 × Zq2 × ... × Zqk

∼ 2 I G = Zp × Zq1 × Zq2 × ... × Zqk

∼ 3 I G = Zp × Zq1 × Zq2 × ... × Zqk

∼ I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk

Automorphism Abelian groups Gerhardt Hinkle

2 Introduction I d(G) = ±p , G abelian Automorphisms Original problem Possibilities: I Generalizations ∼ Prime difference I G = q1 × q2 × ... × qk Z Z Z Prime square I Calculated like in the d(G) = ±p case; can only give difference Possible differences d(G) = −p2 Further research I No solution

∼ 2 I G = Zp3 , d(G) = −p

I k = 1: p = 2, q1 = 5, d(G) = 4 I k = 2: p = 2, q1 = 5, q2 = 7, d(G) = 4 I ... (must be calculated manually)

I Must be calculated manually

∼ 2 I G = Zp × Zq1 × Zq2 × ... × Zqk

∼ 3 I G = Zp × Zq1 × Zq2 × ... × Zqk

∼ I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk

Automorphism Abelian groups Gerhardt Hinkle

∼ 2 I G = Zp3 , d(G) = −p

I k = 1: p = 2, q1 = 5, d(G) = 4 I k = 2: p = 2, q1 = 5, q2 = 7, d(G) = 4 I ... (must be calculated manually)

∼ 2 I G = Zp × Zq1 × Zq2 × ... × Zqk

∼ 3 I G = Zp × Zq1 × Zq2 × ... × Zqk

∼ I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk

Automorphism Abelian groups Gerhardt Hinkle

I k = 1: p = 2, q1 = 5, d(G) = 4 I k = 2: p = 2, q1 = 5, q2 = 7, d(G) = 4 I ... (must be calculated manually)

I No solution

∼ 3 I G = Zp × Zq1 × Zq2 × ... × Zqk

∼ I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk

Automorphism Abelian groups Gerhardt Hinkle

∼ 2 I G = Zp × Zq1 × Zq2 × ... × Zqk ∼ 2 I G = Zp3 , d(G) = −p

I k = 1: p = 2, q1 = 5, d(G) = 4 I k = 2: p = 2, q1 = 5, q2 = 7, d(G) = 4 I ... (must be calculated manually)

∼ 3 I G = Zp × Zq1 × Zq2 × ... × Zqk

∼ I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk

Automorphism Abelian groups Gerhardt Hinkle

∼ 2 I G = Zp × Zq1 × Zq2 × ... × Zqk I No solution I k = 1: p = 2, q1 = 5, d(G) = 4 I k = 2: p = 2, q1 = 5, q2 = 7, d(G) = 4 I ... (must be calculated manually)

∼ 2 I G = Zp3 , d(G) = −p ∼ I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk

Automorphism Abelian groups Gerhardt Hinkle

∼ 2 I G = Zp × Zq1 × Zq2 × ... × Zqk I No solution

∼ 3 I G = Zp × Zq1 × Zq2 × ... × Zqk I k = 1: p = 2, q1 = 5, d(G) = 4 I k = 2: p = 2, q1 = 5, q2 = 7, d(G) = 4 I ... (must be calculated manually)

∼ I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk

Automorphism Abelian groups Gerhardt Hinkle

∼ 2 I G = Zp × Zq1 × Zq2 × ... × Zqk I No solution

∼ 3 I G = Zp × Zq1 × Zq2 × ... × Zqk ∼ 2 I G = Zp3 , d(G) = −p I k = 1: p = 2, q1 = 5, d(G) = 4 I k = 2: p = 2, q1 = 5, q2 = 7, d(G) = 4 I ... (must be calculated manually)

Automorphism Abelian groups Gerhardt Hinkle

∼ 2 I G = Zp × Zq1 × Zq2 × ... × Zqk I No solution

∼ 3 I G = Zp × Zq1 × Zq2 × ... × Zqk ∼ 2 I G = Zp3 , d(G) = −p ∼ I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk I k = 2: p = 2, q1 = 5, q2 = 7, d(G) = 4 I ... (must be calculated manually)

Automorphism Abelian groups Gerhardt Hinkle

∼ 2 I G = Zp × Zq1 × Zq2 × ... × Zqk I No solution

∼ 3 I G = Zp × Zq1 × Zq2 × ... × Zqk ∼ 2 I G = Zp3 , d(G) = −p ∼ I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk I k = 1: p = 2, q1 = 5, d(G) = 4 I ... (must be calculated manually)

Automorphism Abelian groups Gerhardt Hinkle

∼ 2 I G = Zp × Zq1 × Zq2 × ... × Zqk I No solution

∼ 3 I G = Zp × Zq1 × Zq2 × ... × Zqk ∼ 2 I G = Zp3 , d(G) = −p ∼ I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk I k = 1: p = 2, q1 = 5, d(G) = 4 I k = 2: p = 2, q1 = 5, q2 = 7, d(G) = 4 Automorphism Abelian groups Gerhardt Hinkle

∼ 2 I G = Zp × Zq1 × Zq2 × ... × Zqk I No solution

∼ 3 I G = Zp × Zq1 × Zq2 × ... × Zqk ∼ 2 I G = Zp3 , d(G) = −p ∼ I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk I k = 1: p = 2, q1 = 5, d(G) = 4 I k = 2: p = 2, q1 = 5, q2 = 7, d(G) = 4 I ... (must be calculated manually) p p I G/Z(G) =∼ ha, b|a = b = 1, ba = abi p p −1 −1 I In G, a = x, b = y, and bab a = z, where x, y, z ∈ Z(G).

I There may be some elements of Z(G) that are unrelated to any of a, b, x, y, and z, but there can’t be any non-central elements of G that depend on anything but a, b, and elements of Z(G).

Automorphism Non-abelian groups Gerhardt Hinkle

Introduction Automorphisms Original problem Generalizations I Inn(G) =∼ × Prime difference Zp Zp Prime square difference Possible differences Further research p p −1 −1 I In G, a = x, b = y, and bab a = z, where x, y, z ∈ Z(G).

I There may be some elements of Z(G) that are unrelated to any of a, b, x, y, and z, but there can’t be any non-central elements of G that depend on anything but a, b, and elements of Z(G).

Automorphism Non-abelian groups Gerhardt Hinkle

Introduction Automorphisms Original problem Generalizations I Inn(G) =∼ × Prime difference Zp Zp Prime square p p difference I G/Z(G) =∼ ha, b|a = b = 1, ba = abi Possible differences Further research I There may be some elements of Z(G) that are unrelated to any of a, b, x, y, and z, but there can’t be any non-central elements of G that depend on anything but a, b, and elements of Z(G).

Automorphism Non-abelian groups Gerhardt Hinkle

Introduction Automorphisms Original problem Generalizations I Inn(G) =∼ × Prime difference Zp Zp Prime square p p difference I G/Z(G) =∼ ha, b|a = b = 1, ba = abi Possible differences p p −1 −1 Further research I In G, a = x, b = y, and bab a = z, where x, y, z ∈ Z(G). Automorphism Non-abelian groups Gerhardt Hinkle

I There may be some elements of Z(G) that are unrelated to any of a, b, x, y, and z, but there can’t be any non-central elements of G that depend on anything but a, b, and elements of Z(G). I φ(a) = a, φ(b) = bz, φ(z) = z I ψ(a) = az, ψ(b) = b, ψ(z) = z l I χ(a) = ab , χ(b) = b, χ(z) = z

I Let φ, ψ, and χ be automorphisms of G, deﬁned as follows:

I o(φ) = o(ψ) = o(χ) = p, ψ ◦ φ = φ ◦ ψ, χ ◦ φ 6= φ ◦ χ, χ ◦ ψ = ψ ◦ χ ∼ I hφ, ψ, χi = (Zp × Zp) o Zp I The group generated by φ, ψ, and χ is a subgroup of Aut(G). 3 2 I p divides |Aut(G)| and |G|, so |Aut(G)| − |G| = ±p is impossible. I If there are any other elements in Z(G), then G is a direct product of the above group with an abelian group, so p3 still divides |Aut(G)| and |G|.

Automorphism Non-abelian (continued) groups pk pl p Gerhardt Hinkle I G =∼ ha, b, z|a = b = z = 1, ba = abz, za = az, zb = bzi Introduction Automorphisms Original problem Generalizations Prime difference Prime square difference Possible differences Further research I φ(a) = a, φ(b) = bz, φ(z) = z I ψ(a) = az, ψ(b) = b, ψ(z) = z l I χ(a) = ab , χ(b) = b, χ(z) = z I o(φ) = o(ψ) = o(χ) = p, ψ ◦ φ = φ ◦ ψ, χ ◦ φ 6= φ ◦ χ, χ ◦ ψ = ψ ◦ χ ∼ I hφ, ψ, χi = (Zp × Zp) o Zp I The group generated by φ, ψ, and χ is a subgroup of Aut(G). 3 2 I p divides |Aut(G)| and |G|, so |Aut(G)| − |G| = ±p is impossible. I If there are any other elements in Z(G), then G is a direct product of the above group with an abelian group, so p3 still divides |Aut(G)| and |G|.

Automorphism Non-abelian (continued) groups pk pl p Gerhardt Hinkle I G =∼ ha, b, z|a = b = z = 1, ba = abz, za = az, zb = bzi Introduction Automorphisms I Let φ, ψ, and χ be automorphisms of G, defined as Original problem follows: Generalizations Prime difference I φ(a) = a, φ(b) = bz, φ(z) = z Prime square difference I ψ(a) = az, ψ(b) = b, ψ(z) = z Possible differences l I χ(a) = ab , χ(b) = b, χ(z) = z Further research I o(φ) = o(ψ) = o(χ) = p, ψ ◦ φ = φ ◦ ψ, χ ◦ φ 6= φ ◦ χ, χ ◦ ψ = ψ ◦ χ ∼ I hφ, ψ, χi = (Zp × Zp) o Zp I The group generated by φ, ψ, and χ is a subgroup of Aut(G). 3 2 I p divides |Aut(G)| and |G|, so |Aut(G)| − |G| = ±p is impossible. Automorphism Non-abelian (continued) groups pk pl p Gerhardt Hinkle I G =∼ ha, b, z|a = b = z = 1, ba = abz, za = az, zb = bzi Introduction Automorphisms I Let φ, ψ, and χ be automorphisms of G, defined as Original problem follows: Generalizations Prime difference I φ(a) = a, φ(b) = bz, φ(z) = z Prime square difference I ψ(a) = az, ψ(b) = b, ψ(z) = z Possible differences l I χ(a) = ab , χ(b) = b, χ(z) = z Further research I o(φ) = o(ψ) = o(χ) = p, ψ ◦ φ = φ ◦ ψ, χ ◦ φ 6= φ ◦ χ, χ ◦ ψ = ψ ◦ χ ∼ I hφ, ψ, χi = (Zp × Zp) o Zp I The group generated by φ, ψ, and χ is a subgroup of Aut(G). 3 2 I p divides |Aut(G)| and |G|, so |Aut(G)| − |G| = ±p is impossible. I If there are any other elements in Z(G), then G is a direct product of the above group with an abelian group, so p3 still divides |Aut(G)| and |G|. I 10, 26, 34, 50, 52, 58, 86, 100, 116, ...

∼ I e.g. G = Z2 × Z2 × Z385, d(G) = −100

∼ I If G = Zn, then d(G) = φ(n) − n = −(n − φ(n)). I n − φ(n) is called the cototient of n. I Any positive integer that cannot be expressed as n − φ(n) for any positive integer n is called a noncototient.

I The negatives of some noncototients can still be obtained as d(G) for some noncyclic group G.

I If a noncototient equals 2p for some prime p, then I conjecture that d(G) = −2p is impossible.

Automorphism Possible diﬀerences groups Gerhardt Hinkle

∼ I e.g. G = Z2 × Z2 × Z385, d(G) = −100

I n − φ(n) is called the cototient of n. I Any positive integer that cannot be expressed as n − φ(n) for any positive integer n is called a noncototient.

I The negatives of some noncototients can still be obtained as d(G) for some noncyclic group G.

I If a noncototient equals 2p for some prime p, then I conjecture that d(G) = −2p is impossible.

Automorphism Possible diﬀerences groups Gerhardt Hinkle

I All of the groups that were found in the d(G) = ±p Introduction Automorphisms case had d(G) = −p. Therefore, if my conjecture in the Original problem last part of that case is true, then d(G) = p is Generalizations Prime difference impossible. Prime square difference ∼ Possible differences I If G = Zn, then d(G) = φ(n) − n = −(n − φ(n)). Further research I 10, 26, 34, 50, 52, 58, 86, 100, 116, ...

∼ I e.g. G = Z2 × Z2 × Z385, d(G) = −100

I Any positive integer that cannot be expressed as n − φ(n) for any positive integer n is called a noncototient.

I The negatives of some noncototients can still be obtained as d(G) for some noncyclic group G.

I If a noncototient equals 2p for some prime p, then I conjecture that d(G) = −2p is impossible.

Automorphism Possible diﬀerences groups Gerhardt Hinkle

I 10, 26, 34, 50, 52, 58, 86, 100, 116, ... I The negatives of some noncototients can still be obtained as d(G) for some noncyclic group G.

I If a noncototient equals 2p for some prime p, then I conjecture that d(G) = −2p is impossible.

Automorphism Possible diﬀerences groups Gerhardt Hinkle

I The negatives of some noncototients can still be obtained as d(G) for some noncyclic group G.

I If a noncototient equals 2p for some prime p, then I conjecture that d(G) = −2p is impossible.

Automorphism Possible diﬀerences groups Gerhardt Hinkle

I 10, 26, 34, 50, 52, 58, 86, 100, 116, ... ∼ I e.g. G = Z2 × Z2 × Z385, d(G) = −100 I If a noncototient equals 2p for some prime p, then I conjecture that d(G) = −2p is impossible.

Automorphism Possible diﬀerences groups Gerhardt Hinkle

I 10, 26, 34, 50, 52, 58, 86, 100, 116, ... I The negatives of some noncototients can still be obtained as d(G) for some noncyclic group G. I If a noncototient equals 2p for some prime p, then I conjecture that d(G) = −2p is impossible.

Automorphism Possible diﬀerences groups Gerhardt Hinkle

I 10, 26, 34, 50, 52, 58, 86, 100, 116, ... I The negatives of some noncototients can still be obtained as d(G) for some noncyclic group G. ∼ I e.g. G = Z2 × Z2 × Z385, d(G) = −100 Automorphism Possible diﬀerences groups Gerhardt Hinkle

I 10, 26, 34, 50, 52, 58, 86, 100, 116, ... I The negatives of some noncototients can still be obtained as d(G) for some noncyclic group G. ∼ I e.g. G = Z2 × Z2 × Z385, d(G) = −100 I If a noncototient equals 2p for some prime p, then I conjecture that d(G) = −2p is impossible. I |Inn(G)| = 1, 2, p, 2p ∼ I Inn(G) = {e}, D2p ∼ I Either G is abelian or G/Z(G) = D2p.

Automorphism Noncototient diﬀerence groups Gerhardt Hinkle

Introduction Automorphisms Original problem Generalizations Prime difference Prime square difference I d(G) = |Aut(G)| − |G| = −2p, where 2p is a Possible differences noncototient Further research ∼ I Inn(G) = {e}, D2p ∼ I Either G is abelian or G/Z(G) = D2p.

Automorphism Noncototient diﬀerence groups Gerhardt Hinkle

I |Inn(G)| = 1, 2, p, 2p ∼ I Either G is abelian or G/Z(G) = D2p.

Automorphism Noncototient diﬀerence groups Gerhardt Hinkle

I |Inn(G)| = 1, 2, p, 2p ∼ I Inn(G) = {e}, D2p Automorphism Noncototient diﬀerence groups Gerhardt Hinkle

I |Inn(G)| = 1, 2, p, 2p ∼ I Inn(G) = {e}, D2p ∼ I Either G is abelian or G/Z(G) = D2p. I 6(q1 − 1)(q2 − 1)...(qk − 1) − 4q1q2...qk = −2p I 3(q1 − 1)(q2 − 1)...(qk − 1) − 2q1q2...qk = −p I The left side is even but the right side is odd, so there is no solution.

I This case has the same problem as the previous case, so there is no solution. ∼ I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk

∼ I G = Z2 × Z2 × Zq1 × Zq2 × ... × Zqk

∼ I G = Z2 × Z2 × Zp × Zp × Zq1 × Zq2 × ... × Zqk

Automorphism Abelian groups Gerhardt Hinkle

Introduction Automorphisms Original problem Generalizations I Possible cases: Prime difference Prime square difference Possible differences Further research I This case has the same problem as the previous case, so there is no solution. ∼ I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk

I 6(q1 − 1)(q2 − 1)...(qk − 1) − 4q1q2...qk = −2p I 3(q1 − 1)(q2 − 1)...(qk − 1) − 2q1q2...qk = −p I The left side is even but the right side is odd, so there is no solution. ∼ I G = Z2 × Z2 × Zp × Zp × Zq1 × Zq2 × ... × Zqk

Automorphism Abelian groups Gerhardt Hinkle

Introduction Automorphisms Original problem Generalizations I Possible cases: Prime difference ∼ Prime square I G = 2 × 2 × q × q × ... × q difference Z Z Z 1 Z 2 Z k Possible differences Further research I This case has the same problem as the previous case, so there is no solution. ∼ I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk

I 3(q1 − 1)(q2 − 1)...(qk − 1) − 2q1q2...qk = −p I The left side is even but the right side is odd, so there is no solution. ∼ I G = Z2 × Z2 × Zp × Zp × Zq1 × Zq2 × ... × Zqk

Automorphism Abelian groups Gerhardt Hinkle

Introduction Automorphisms Original problem Generalizations I Possible cases: Prime difference ∼ Prime square I G = 2 × 2 × q × q × ... × q difference Z Z Z 1 Z 2 Z k Possible differences I 6(q1 − 1)(q2 − 1)...(qk − 1) − 4q1q2...qk = −2p Further research I This case has the same problem as the previous case, so there is no solution. ∼ I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk

I The left side is even but the right side is odd, so there is no solution. ∼ I G = Z2 × Z2 × Zp × Zp × Zq1 × Zq2 × ... × Zqk

Automorphism Abelian groups Gerhardt Hinkle

∼ I G = Z2 × Z2 × Zp × Zp × Zq1 × Zq2 × ... × Zqk

Automorphism Abelian groups Gerhardt Hinkle

Introduction Automorphisms Original problem Generalizations I Possible cases: Prime difference ∼ Prime square I G = 2 × 2 × q × q × ... × q difference Z Z Z 1 Z 2 Z k Possible differences I 6(q1 − 1)(q2 − 1)...(qk − 1) − 4q1q2...qk = −2p Further research I 3(q1 − 1)(q2 − 1)...(qk − 1) − 2q1q2...qk = −p I The left side is even but the right side is odd, so there is no solution. I This case has the same problem as the previous case, so there is no solution. ∼ I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk

Automorphism Abelian groups Gerhardt Hinkle

Introduction Automorphisms Original problem Generalizations I Possible cases: Prime difference ∼ Prime square I G = 2 × 2 × q × q × ... × q difference Z Z Z 1 Z 2 Z k Possible differences I 6(q1 − 1)(q2 − 1)...(qk − 1) − 4q1q2...qk = −2p Further research I 3(q1 − 1)(q2 − 1)...(qk − 1) − 2q1q2...qk = −p I The left side is even but the right side is odd, so there is no solution. ∼ I G = Z2 × Z2 × Zp × Zp × Zq1 × Zq2 × ... × Zqk I This case has the same problem as the previous case, so there is no solution. Automorphism Abelian groups Gerhardt Hinkle

Introduction Automorphisms Original problem Generalizations I Possible cases: Prime difference ∼ Prime square I G = 2 × 2 × q × q × ... × q difference Z Z Z 1 Z 2 Z k Possible differences I 6(q1 − 1)(q2 − 1)...(qk − 1) − 4q1q2...qk = −2p Further research I 3(q1 − 1)(q2 − 1)...(qk − 1) − 2q1q2...qk = −p I The left side is even but the right side is odd, so there is no solution. ∼ I G = Z2 × Z2 × Zp × Zp × Zq1 × Zq2 × ... × Zqk I This case has the same problem as the previous case, so there is no solution. ∼ I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk 2 2 2 I (p −1)(p −p)(q1 −1)(q2 −1)...(qk −1)−p q1q2...qk = −2p 2 I (p −1)(p−1)(q1−1)(q2−1)...(qk −1)−pq1q2...qk = −2

I The left side is odd unless one of the qi s is 2, so let q1 = 2. 2 I (p −1)(p −1)(q2 −1)(q3 −1)...(qk −1)−2pq2q3...qk = −2

I ri = qi+1 2 I (p −1)(p−1)(r1−1)(r2−1)...(rk−1−1)−2pr1r2...rk−1 = −2

I Using a similar argument as in the last case for d(G) = ±p, the lower bound on k − 1 is k(p).

I Therefore, I conjecture that if 2p is a noncototient, then there are no abelian groups G for which d(G) = −2p.

Automorphism Abelian (continued) groups ∼ Gerhardt Hinkle I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk Introduction Automorphisms Original problem Generalizations Prime difference Prime square difference Possible differences Further research 2 I (p −1)(p−1)(q1−1)(q2−1)...(qk −1)−pq1q2...qk = −2

I The left side is odd unless one of the qi s is 2, so let q1 = 2. 2 I (p −1)(p −1)(q2 −1)(q3 −1)...(qk −1)−2pq2q3...qk = −2

I ri = qi+1 2 I (p −1)(p−1)(r1−1)(r2−1)...(rk−1−1)−2pr1r2...rk−1 = −2

I Using a similar argument as in the last case for d(G) = ±p, the lower bound on k − 1 is k(p).

I Therefore, I conjecture that if 2p is a noncototient, then there are no abelian groups G for which d(G) = −2p.

Automorphism Abelian (continued) groups ∼ Gerhardt Hinkle I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk Introduction 2 2 2 I (p −1)(p −p)(q1 −1)(q2 −1)...(qk −1)−p q1q2...qk = Automorphisms Original problem −2p Generalizations Prime difference Prime square difference Possible differences Further research I The left side is odd unless one of the qi s is 2, so let q1 = 2. 2 I (p −1)(p −1)(q2 −1)(q3 −1)...(qk −1)−2pq2q3...qk = −2

I ri = qi+1 2 I (p −1)(p−1)(r1−1)(r2−1)...(rk−1−1)−2pr1r2...rk−1 = −2

I Using a similar argument as in the last case for d(G) = ±p, the lower bound on k − 1 is k(p).

I Therefore, I conjecture that if 2p is a noncototient, then there are no abelian groups G for which d(G) = −2p.

Automorphism Abelian (continued) groups ∼ Gerhardt Hinkle I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk Introduction 2 2 2 I (p −1)(p −p)(q1 −1)(q2 −1)...(qk −1)−p q1q2...qk = Automorphisms Original problem −2p Generalizations 2 Prime difference I (p −1)(p−1)(q1−1)(q2−1)...(qk −1)−pq1q2...qk = −2 Prime square difference Possible differences Further research 2 I (p −1)(p −1)(q2 −1)(q3 −1)...(qk −1)−2pq2q3...qk = −2

I ri = qi+1 2 I (p −1)(p−1)(r1−1)(r2−1)...(rk−1−1)−2pr1r2...rk−1 = −2

I Using a similar argument as in the last case for d(G) = ±p, the lower bound on k − 1 is k(p).

I Therefore, I conjecture that if 2p is a noncototient, then there are no abelian groups G for which d(G) = −2p.

Automorphism Abelian (continued) groups ∼ Gerhardt Hinkle I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk Introduction 2 2 2 I (p −1)(p −p)(q1 −1)(q2 −1)...(qk −1)−p q1q2...qk = Automorphisms Original problem −2p Generalizations 2 Prime difference I (p −1)(p−1)(q1−1)(q2−1)...(qk −1)−pq1q2...qk = −2 Prime square difference Possible differences I The left side is odd unless one of the qi s is 2, so let Further research q1 = 2. I ri = qi+1 2 I (p −1)(p−1)(r1−1)(r2−1)...(rk−1−1)−2pr1r2...rk−1 = −2

I Using a similar argument as in the last case for d(G) = ±p, the lower bound on k − 1 is k(p).

I Therefore, I conjecture that if 2p is a noncototient, then there are no abelian groups G for which d(G) = −2p.

Automorphism Abelian (continued) groups ∼ Gerhardt Hinkle I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk Introduction 2 2 2 I (p −1)(p −p)(q1 −1)(q2 −1)...(qk −1)−p q1q2...qk = Automorphisms Original problem −2p Generalizations 2 Prime difference I (p −1)(p−1)(q1−1)(q2−1)...(qk −1)−pq1q2...qk = −2 Prime square difference Possible differences I The left side is odd unless one of the qi s is 2, so let Further research q1 = 2. 2 I (p −1)(p −1)(q2 −1)(q3 −1)...(qk −1)−2pq2q3...qk = −2 2 I (p −1)(p−1)(r1−1)(r2−1)...(rk−1−1)−2pr1r2...rk−1 = −2

I Using a similar argument as in the last case for d(G) = ±p, the lower bound on k − 1 is k(p).

I Therefore, I conjecture that if 2p is a noncototient, then there are no abelian groups G for which d(G) = −2p.

Automorphism Abelian (continued) groups ∼ Gerhardt Hinkle I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk Introduction 2 2 2 I (p −1)(p −p)(q1 −1)(q2 −1)...(qk −1)−p q1q2...qk = Automorphisms Original problem −2p Generalizations 2 Prime difference I (p −1)(p−1)(q1−1)(q2−1)...(qk −1)−pq1q2...qk = −2 Prime square difference Possible differences I The left side is odd unless one of the qi s is 2, so let Further research q1 = 2. 2 I (p −1)(p −1)(q2 −1)(q3 −1)...(qk −1)−2pq2q3...qk = −2

I ri = qi+1 I Using a similar argument as in the last case for d(G) = ±p, the lower bound on k − 1 is k(p).

I Therefore, I conjecture that if 2p is a noncototient, then there are no abelian groups G for which d(G) = −2p.

Automorphism Abelian (continued) groups ∼ Gerhardt Hinkle I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk Introduction 2 2 2 I (p −1)(p −p)(q1 −1)(q2 −1)...(qk −1)−p q1q2...qk = Automorphisms Original problem −2p Generalizations 2 Prime difference I (p −1)(p−1)(q1−1)(q2−1)...(qk −1)−pq1q2...qk = −2 Prime square difference Possible differences I The left side is odd unless one of the qi s is 2, so let Further research q1 = 2. 2 I (p −1)(p −1)(q2 −1)(q3 −1)...(qk −1)−2pq2q3...qk = −2

I ri = qi+1 2 I (p −1)(p−1)(r1−1)(r2−1)...(rk−1−1)−2pr1r2...rk−1 = −2 I Therefore, I conjecture that if 2p is a noncototient, then there are no abelian groups G for which d(G) = −2p.

Automorphism Abelian (continued) groups ∼ Gerhardt Hinkle I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk Introduction 2 2 2 I (p −1)(p −p)(q1 −1)(q2 −1)...(qk −1)−p q1q2...qk = Automorphisms Original problem −2p Generalizations 2 Prime difference I (p −1)(p−1)(q1−1)(q2−1)...(qk −1)−pq1q2...qk = −2 Prime square difference Possible differences I The left side is odd unless one of the qi s is 2, so let Further research q1 = 2. 2 I (p −1)(p −1)(q2 −1)(q3 −1)...(qk −1)−2pq2q3...qk = −2

I ri = qi+1 2 I (p −1)(p−1)(r1−1)(r2−1)...(rk−1−1)−2pr1r2...rk−1 = −2

I Using a similar argument as in the last case for d(G) = ±p, the lower bound on k − 1 is k(p). Automorphism Abelian (continued) groups ∼ Gerhardt Hinkle I G = Zp × Zp × Zq1 × Zq2 × ... × Zqk Introduction 2 2 2 I (p −1)(p −p)(q1 −1)(q2 −1)...(qk −1)−p q1q2...qk = Automorphisms Original problem −2p Generalizations 2 Prime difference I (p −1)(p−1)(q1−1)(q2−1)...(qk −1)−pq1q2...qk = −2 Prime square difference Possible differences I The left side is odd unless one of the qi s is 2, so let Further research q1 = 2. 2 I (p −1)(p −1)(q2 −1)(q3 −1)...(qk −1)−2pq2q3...qk = −2

I ri = qi+1 2 I (p −1)(p−1)(r1−1)(r2−1)...(rk−1−1)−2pr1r2...rk−1 = −2

I Using a similar argument as in the last case for d(G) = ±p, the lower bound on k − 1 is k(p).

I Therefore, I conjecture that if 2p is a noncototient, then there are no abelian groups G for which d(G) = −2p. I Finish the abelian case for d(G) = −2p when 2p is a noncototient

I Do the non-abelian case for d(G) = −2p when 2p is a ∼ noncototient (G/Z(G) = D2p) n I Extend to d(G) = ±p , d(G) = ±pq, etc.

I Determine what other diﬀerences are possible or impossible

Automorphism Further research groups Gerhardt Hinkle

Introduction Automorphisms Original problem Generalizations I Finish the last case for d(G) = ±p Prime difference Prime square difference Possible differences Further research I Do the non-abelian case for d(G) = −2p when 2p is a ∼ noncototient (G/Z(G) = D2p) n I Extend to d(G) = ±p , d(G) = ±pq, etc.

I Determine what other diﬀerences are possible or impossible

Automorphism Further research groups Gerhardt Hinkle

Introduction Automorphisms Original problem Generalizations I Finish the last case for d(G) = ±p Prime difference Prime square difference I Finish the abelian case for d(G) = −2p when 2p is a Possible differences noncototient Further research n I Extend to d(G) = ±p , d(G) = ±pq, etc.

I Determine what other diﬀerences are possible or impossible

Automorphism Further research groups Gerhardt Hinkle

I Do the non-abelian case for d(G) = −2p when 2p is a ∼ noncototient (G/Z(G) = D2p) I Determine what other diﬀerences are possible or impossible

Automorphism Further research groups Gerhardt Hinkle

I Do the non-abelian case for d(G) = −2p when 2p is a ∼ noncototient (G/Z(G) = D2p) n I Extend to d(G) = ±p , d(G) = ±pq, etc. Automorphism Further research groups Gerhardt Hinkle

I Do the non-abelian case for d(G) = −2p when 2p is a ∼ noncototient (G/Z(G) = D2p) n I Extend to d(G) = ±p , d(G) = ±pq, etc.

I Determine what other diﬀerences are possible or impossible