Exercise. Describe All Homomorphisms from Z24 to Z18. Solution

Home , Group homomorphism, Homomorphism

Exercise. Describe all homomorphisms from Z24 to Z18. Solution. We need some preliminary discussion before we try to define the homomorphisms to try to figure out how they might look like. Let f : Z24 → Z18 be a homomorphism. For any integer k ≥ 0 notice that f(k mod 24) = f((1 mod 24)+ ··· +(1mod 24)) (added k times) = f(1 mod 24)+ ··· + f(1 mod 24) = k · f(1 mod 24). It follows that a homomorphism f is completely determined by the value f(1 mod 24). Write f(1 mod 24) = n mod 18 where n is an integer such that 0 ≤ n ≤ 17. Notice that 24n mod 18 = 24f(1 mod 24) = f(24 mod 24) = f(0 mod 24) = 0 mod 18. since any homomorphism maps the identity to identity. So 18 must divide 24n, so 3 must divide 4n, hence 3 must divide n. So n must be among 0, 3, 6, 9, 12, 15. The above discussion suggests how we might want to define our homomorphisms. We find that any such homomorphism must send (1 mod 24) to (3k mod 18) for some k =0, 1, ··· , 5. It follows that there are at most 6 possible homomorphisms from Z24 → Z18, given by f0, f1, ··· , f5 where fk(r mod 24) = 3kr mod18 (1) for k =0, 1, ··· , 5. First we need to check that for each of this k, equation (1) gives a well ′ ′ defined function from Z24 to Z18. To this end, let r, r ∈ Z such that r mod 24 = r mod 24. Then (r − r′) is a multiple of 24. So (3kr − 3kr′)=3k(r − r′) is a multiple of 3 × 24 = 72. In particular, (3kr − 3kr′) is a multiple of 18. It follows that (3kr mod 18) only depends on the congruence class of r modulo 24. So indeed equation (1) gives us a well defined function from Z24 to Z18. Now we verify that for any r, s ∈ Z, we have

fk(r mod 24+ s mod 24) = fk((r + s)mod 24)= 3k(r + s) mod 18 =3kr mod 18+3ks mod 18

= fk(r mod 18) + fk(s mod 18).

So each fk is indeed a homomorphism from Z24 to Z18. At the risk of repeating ourselves, we recall that if f : Z24 → Z18 is any homomorphism, then f(1 mod 18) must be of the form (3k mod 18) for some k = 0, 1, ··· , 5 and that any homomorphism f is determined completely by the value of f(1 mod 18). For k =0, 1, ··· , 5, we have constructed a homomorphism fk such that fk(1mod18) = 3k mod 18. It follows that there are six homomorphisms from Z24 to Z18 and these are given by f0, f1, ··· , f5 as deﬁned in equation (1).

1 2. If m is an integer, then mZ denotes the set of all integer multiples of m. 2.1. Lemma. Let m, n be nonzero integers and r = lcm(m, n). Then mZ ∩ nZ = rZ. Proof. Let x ∈ mZ ∩ nZ. Then x is a multiple of m and x is a multiple of n. Hence x is a multiple of r. So x ∈ rZ. So mZ ∩ nZ ⊆ rZ. On the other hand r is a multiple of m and n, so r ∈ mZ ∩ nZ. So rZ ⊆ mZ ∩ nZ since mZ ∩ nZ is a subgroup of Z. 2.2. Lemma. Let G be a group and a be an element of G. Then hami = {ak : k ∈ mZ}. Proof. If k ∈ mZ, then k has the form k = mt for some integer t, so ak = amt ∈hami. Thus {ak : k ∈ mZ}⊆hami. On the other hand, if g ∈ hami, then g has the form g = ams for some integer s. Let k = ms. Then g = ak and k ∈ mZ. So g ∈{ak : k ∈ mZ}. 2.3. Lemma. Let G be a group and a ∈ G. Then show that hami∩hani = halcm(m,n)i. Proof. Write r = lcm(m, n). From the previous lemma, we have hami∩hani = {ak : k ∈ mZ}∩{ak : k ∈ nZ} = {ak : k ∈ mZ ∩ nZ} = {ak : k ∈ rZ} = hari. Here the first and the final equality follows from lemma 2.2 and the third equality follows from lemma 2.1. The above is a rather straight forward proof. On the next page you will find a more elaborate argument which also ends up giving a proof of the same statement in a much more long winded manner. But along the way we prove a bunch of things which are of some importance in themselves.

2 All through the following discussion, let G be a group and let a be an element of G of finite order |a| = d. If m is an integer, we shall write m′ = gcd(m, d). Recall from the text that |am| = d/m′. 2.4. Lemma. If m and n are relatively prime nonzero integers, then (mn)′ = m′n′. Proof. Note that m′ is a factor of m and n′ is a factor of n. Since m and n are relatively prime, so are m′ and n′. Since m′ and n′ each divide d, it follows that m′n′ divides d. Since (m/m′) and (n/n′) are each relatively prime to d, so is the product (m/m′)(n/n′). Writing mn =(m′n′)((m/m′)(n/n′)), we find that gcd(mn, d)= m′n′. 2.5. Lemma. Let k, n be nonzero integers. If k is a multiple of n, then haki⊆hani. Proof. We can write k = nt for some t ∈ Z. So ak = ant ∈hani. So haki⊆hani. ′ 2.6. Lemma. Let m is a nonzero integer, then hami = ham i. ′ Proof. Note that m is a multiple of m′, so lemma 2.5 implies am ∈ham i. Conversely, we can write m′ = αm + β|a| for some integers α, β ∈ Z. So ′ am = aαm+β|a| = aαm ∈hami. 2.7. Lemma. (a) If b ∈hai, then |b| divides |a|. (b)If k, m are nonzero integers such that ak ∈hami, then m′ | k′. Proof. (a) We can write b = at for some nonzero integer t. So |b| = |a|/t′ is a factor of |a|. (b) By part (a), we have |ak| divides |am|, that is, |a|/k′ divides |a|/m′. So m′ divides k′. 2.8. Lemma. Let m, n be relatively prime nonzero integers. Then hami∩hani = hamni. Proof. Write H = hami∩hani. By lemma 2.5, we have hamni⊆hami and hamni⊆hani. So hamni ⊆ H. Notice that H is a subgroup of the cyclic group hami, so H is cyclic. So H = haki for some nonzero integer k. Since ak ∈ hami and ak ∈ hani, lemma 2.7(b) implies that m′ | k′ and that n′ | k′. Since m′ and n′ are relatively prime, we have m′n′ | k′. So k′ is a multiple ′ ′ of m′n′ =(mn)′ (by lemma 2.4) and lemma 2.5 implies that hak i⊆ha(mn) i. Now we obtain ′ ′ H = haki = hak i⊆ha(mn) i = hamni where the second and the final equality follows from lemma 2.6. 2.9. Lemma. Let m and n be two nonzero integers and l = lcm(m, n). Then hami∩hani = hali. Proof. Since a−m and am generate the same subgroup, without loss, we may assume that m is a positive integer and similarly, that n is a positive integer. Let d = gcd(m, n). Let b = ad. Then am = bm/d and an = bn/d. Now we have 2 hami∩hani = hbm/di∩hbn/di = hbmn/d i = hamn/di = hali where the second equality follows from 2.8 since (m/d) and (n/d) are relatively prime and the last equality follows since mn = dl. 3 3. Sign of a permutation, even and odd permutations. 3.1. Definition. We want to define the “sign” of a permutation. We start with the polynomial P (x1, ··· , xn) in n variables x1, ··· , xn with n(n − 1)/2 distinct linear factors (xi − xj) for 1 ≤ i < j ≤ n. For example, if n = 4, then

P (x1, x2, x3, x4)=(x1 − x2)(x1 − x3)(x1 − x4)(x2 − x3)(x2 − x4)(x3 − x4).

Let α ∈ Sn be any permutation. Say that a pair (i, j) is a crossing pair for α if (i < j) −1 −1 1 2 ··· n and (α >α (j)), that is, when we represent α in the form α(1) α(2) ······ α(n) then in the second row “i is sitting under a bigger number than j”, that is, “i is sitting to the right of j”. Let N(α) be set of crossing pairs for α N(α)= {(i, j): 1 ≤ i < j ≤ n, α−1(i) >α−1(j)}. The number |N(α)| is called the number of crossings of α.

Claim: The polynomial P (xα(1), ··· , xα(n)) also has n(n − 1)/2 distinct linear factors. Upto a plus or minus sign, these factors are same as the factors of P (x1, ··· , xn). More precisely, for each 1 ≤ i < j ≤ n, either (xi − xj) or (xj − xi) is a factor of P (xα(1), ··· , xα(n)), but not both. If (i, j) ∈/ N(α), then (xi − xj) is a factor, otherwise (xj − xi) is a factor. proof of claim: Pick a pair i, j such that 1 ≤ i < j ≤ n. Suppose (i, j) ∈/ N(α). Then −1 −1 α (i) <α (j), so (xα−1(i) −xα−1(j)) is a factor of P (x1, ··· , xn). So(xα(α−1(i)) −xα(α−1 (j)))= (xi − xj) is a factor of P (xα(1), ··· , xα(n)). On the other hand, suppose (i, j) ∈ N(α). Then −1 −1 α (j) <α (i), so (xα−1(j) −xα−1(i)) is a factor of P (x1, ··· , xn). So(xα(α−1(j)) −xα(α−1(i)))= (xj − xi) is a factor of P (xα(1), ··· , xα(n)). Thus we ﬁnd that for each i, j such that 1 ≤ i < j ≤ n, either (xi − xj) or (xj − xi) is a factor of P (xα(1), ··· , xα(n)). This already gives n(n − 1)/2 distinct factors which is equal to the degree of the polynomial, so each of these factors must occur just once and there cannot be any more factors.

1234 −1 1234 3.2. Example. Let a = 2413 . So a = 3142 . Then N(a)= {(1, 2), (1, 4), (3, 4)} and P (xa(1), xa(2), xa(3), xa(4))=(x2 − x4)(x2 − x1)(x2 − x3)(x4 − x1)(x4 − x3)(x1 − x3). Observe that we get a contribution of −1 for each crossing pair (i, j).

From the claim above we ﬁnd that the ratio P (xα(1), ··· , xα(n))/P (x1, ··· , xn) is equal to either 1 or −1. Deﬁne P (x , ··· , x ) sign(α)= α(1) α(n) . P (x1, ··· , xn) The number sign(α) is called the sign of the permutation α.

3.3. Theorem. If α, β ∈ Sn, then sign(αβ) = sign(α) sign(β), that is, the sign is multiplicative. In other words, the function sign : Sn →{1, −1} is a group homomorphism. Proof. By definition of sign of a permutation, we have P (x , ··· , x ) P (x , ··· , x ) P (x , ··· , x ) sign(αβ)= α(β(1)) α(β(n)) = α(1) α(n) · α(β(1)) α(β(n)) (2) P (x1, ··· , xn) P (x1, ··· , xn) P (xα(1), ··· , xα(n)) Notice that the first fraction in the right hand side of (2) is equal to sign(α). Introduce the variables y1 = xα(1), ··· ,yn = xα(n). Then yβ(1) = xα(β(1)), ··· ,yβ(1) = xα(β(1)). Writing the 4 second fraction in the right hand side of (2) in terms of the variables y1, ··· ,yn, we get P (x , ··· , x ) P (y , ··· ,y ) α(β(1)) α(β(n)) = β(1) β(n) = sign(β). P (xα(1), ··· , xα(n)) P (y1, ··· ,yn) Now equation (2) implies that sign(αβ) = sign(α) sign(β). 3.4. Remark. Note that sign(id) = 1. Further, since sign of a permutation is either 1 or −1, we have sign(α−1) = sign(α)−1 = sign(α) and sign(αβα−1) = sign(α) sign(β) sign(α)−1 = sign(β). 3.5. Lemma. The sign of a flip (or a transposition) is −1. Proof. Directly verify from the definition that sign(1 2) = −1 (left as an exercise). Let (i j) be any flip. Then Pick a permutation α such that α(1) = i and α(2) = j. Then (i j)= α(1 2)α−1. So sign(i j) = sign(α) sign(12) sign(α)−1 = sign(1 2) = −1. 3.6. Theorem. (a) If a permutation can be written as a product of a even number of flips, then it has sign 1. (b) If a permutation can be written as a product of a odd number of flips, then it has sign −1. (c) If a permutation can be written as a product of a even number of flips, then it cannot be written as a product of an odd number of flips and vice versa.

Proof. Let α be a permutation. Suppose α = β1 ··· β2k where each βj is a flip. Then 2k sign(α) = sign(β1) ··· sign(β2k)=(−1) =1. Note that the first equality above holds since sign is multiplicative (see theorem 3.3) and the second equality holds by lemma 3.5. This proves part (a). The proof of (b) is exactly similar. Part (c) follows from part (a) and (b) since a permutation has a well defined sign. The theorem above prompts the next definition. 3.7. Definition. A permutation is called even if it can be written as a product of even number of flips, otherwise the permutation is called odd. Notice that the notion of even and odd permutations make sense because of part (c) of the theorem 3.6. Also notice that an even permutation is just another name for a permutation that have sign 1 and that an odd permutation is just another name for a permutation that have sign −1. We may restate some of the above facts about sign of a permutation, in terms of even and odd permutations as follows. These facts follow directly from the corresponding facts about sign of a permutation. ◦ The identity is an even permutation while each flip is odd. ◦ Since the sign of a permutation is multiplicative, the product of two even permutations is even and the product of two odd permutations is even. The product of an even and an odd permutation is odd. The inverse of an even permutation is even and the inverse of an odd permutation is odd. ◦ In particular, the set of even permutations form a subgroup of Sn. This subgroup is called the alternating group and is denoted by An. 3.8. Exercise: Show that sgn(α)=(−1)|N(α)|. (Hint: Look at the proof of the claim in 3.1)

5 4. Some exercises: November 11. 4.1. Exercise. Prove or disprove: There is a non-cyclic abelian group of order 20.

Solution. Let G = Z2 × Z10. Let g1,g2 ∈ G. We can write these elements in the form g1 =(α1, β1) and g2 =(α2, β2) for some α1,α2 ∈ Z2 and β1, β2 ∈ Z10. The composition law on G is deﬁned by g1 + g2 =(α1 + β1,α2 + β2)

It follows that g1 + g2 = g2 + g1 for all g1,g2 ∈ G, so G is abelian. Note that |G| = |Z2||Z10| = 20. Let g ∈ G. Then g has the form g =(a mod 2, b mod 10) for some a, b ∈ Z. So 10g = (10a mod 2, 10b mod 10) = (0 mod 2, 0 mod 10) and this is the identity of G. So every element of G has order at most 10. In particular G does not have any element of order 20, so G is not cyclic. So there is a non-cyclic abelian group of order 20. −1 4.2. Exercise. Let φ : G → H be a group homomorphism. Show that φ {eH } = {eG} if and only if φ is one to one. Here eG and eH denotes the identities of G and H respectively. −1 −1 Also recall that φ {eH } is deﬁned as the set φ {eH } = {g ∈ G: φ(g)= eH }.

Solution. Suppose φ is one to one. Since φ is a homomorphism, we know that φ(eG) = eH . −1 Since φ is one to one, more than one element in G cannot map to eH , so φ {eH } can have −1 at most one element. It follows that φ {eH } = {eG}. −1 Conversely, suppose φ {eH } = {eG}. Let a, b ∈ G such that φ(a)= φ(b). Then −1 −1 −1 φ(ab )= φ(a)φ(b) = φ(a)φ(a) = eH . −1 −1 −1 So ab ∈ φ {eH } = {eG}. In other words, ab = eG, so a = b. Hence φ is one to one. 4.3. Exercise. Let G be a group. Let a, b ∈ G such that |a| = m and |b| = n. If gcd(m, n)= 1, then show that hai∩hbi = {e}. Solution. Let e denote the identity of G. Since gcd(m, n) = 1, there exists integers r and s such that mr + ns = 1. Let x ∈hai∩hbi. Since x ∈hai we can write x = at for some t ∈ Z and hence xm = atm =(am)t = et = e. Similarly, since x ∈hbi and |b| = n, we have xn = e. So x = xmr+ns =(xm)r(xn)s = eres = e.

4.4. Exercise. (a) Calculate the order of (6 mod 17) in U(17). Show that U(17) is a cyclic group with generator (6 mod 17) (b) What is the remainder obtained when 142491 is divided by 17? You may use the fact that 2491 = 155.16+11. Proof. (a) We calculate 62 ≡ 2mod17. So 64 ≡ 4mod17 and 68 ≡ 16 ≡ −1 mod 17 and 616 ≡ 1 mod 17. Write a = 6mod17. So a16 = 1 mod 17, hence the order of a is a factor of 16. But we already saw that an =6 1 mod 17 for n = 1, 2, 4, 8. It follows that the order of a in U(17) is 16. Since 17 is a prime, we have U(17) = {n mod 17: 1 ≤ n ≤ 16, n ∈ Z}. So U(17) has order 16. Since we found an element of order 16 in U(17), it follows that U(17) is cyclic and a is a generator for this cyclic group. 6 (b) Since U(17) = hai, we can write (14 mod 17) = ak for some integer k. Now a2491 = a155.16+11 = a11 since a16 = 1mod17. So (14 mod 17)2491 = a2491k = a11k = (14 mod 17)11. Now we have 142 ≡ (−3)2 ≡ 9 mod 17, 144 ≡ 81 ≡ (−4) mod 17, 148 ≡ 16 ≡ −1 mod 17. So 1411 = 148.142.14 ≡ (−1).9.(−3) ≡ 27 ≡ 10 mod 17. So 142491 ≡ 10 mod 17. So the remainder is 10. 157 4.5. Exercise. Find an element a of largest order in S8. What is the order of a? Find a .

Proof. Let n1 ≥ n2 ≥ ··· ≥ nk be a non-increasing sequence of natural numbers with n1 + n2 + ···+ nk = 8. We shall say that a permutation σ ∈ S8 has cycle type (n1, ··· , nk) if σ, when written as a product of disjoint cycles, consists of k cycles of length n1, n2, ··· , nk respectively. For example, the permutation (1468)(253)(7) has cycle type (4, 3, 1). Note that if a permutation has cycle type (n1, ··· , nk), then its order is the least common multiple of the numbers n1, ··· , nk. We can list all the cycle types in S8 by listing all nonincreasing sequences of natural numbers (n1, ··· , nk) such that Pk nk = 8. There are 22 such cycle types. They are (8), (7, 1), (6, 2), (6, 1, 1)(5, 3), (5, 2, 1), (5, 1, 1, 1), (4, 4), (4, 3, 1), (4, 2, 2), (4, 2, 1, 1), (4, 1, 1, 1, 1) (3, 3, 2), (3, 3, 1, 1), (3, 2, 2, 1), (3, 2, 1, 1, 1), (3, 1, 1, 1, 1, 1) (2, 2, 2, 2), (2, 2, 2, 1, 1), (2, 2, 1, 1, 1, 1), (2, 1, 1, 1, 1, 1, 1)(1, 1, 1, 1, 1, 1, 1, 1). Looking at the least common multiples, we ﬁnd that the largest order occurs for cycle type (5, 3). So a = (1 2 3 4 5)(6 7 8) is an element of largest order in S8. One has |a| = 15. So a15 = e where e denotes the identity permutation. So a157 = (a15)10a7 = a7. Write b = (1 2 3 4 5) and c = (6 7 8). The cycles b and c are disjoint, so they commute, that is, bc = cb. So a7 = (bc)7 = b7c7. Since b has order 5 and c has order 3, we have b7 = b2 =(13524) and c7 = c = (6 7 8). Finally, we get a157 = a7 = b7c7 = b2c =(13524)(678).

4.6. Exercise. Let φ : G → H be a group homomorphism. Let N = {g ∈ G: φ(g) = eH }. Show that N is a subgroup of G.

Solution. Since φ is a group homomorphism, we have φ(eG)= eH , so eG ∈ N.

Let a, b ∈ N. Then φ(a) = eH and φ(b) = eH . Since φ is a homomorphism, we have φ(ab)= φ(a)φ(b)= eH .eH = eH . So ab ∈ N if a and b belong to N.

−1 Finally, let a ∈ N. Then φ(a) = eH . Since φ is a homomorphism, we have φ(a ) = −1 −1 −1 (φ(a)) = eH = eH . So a ∈ N if a ∈ N.

It follows that N is a subgroup of G.

7 5. normal subgroup and quotient group We begin by stating a couple of elementary lemmas. 5.1. Lemma. Let A and B be subsets of a group G. If g, h ∈ G, then g(hA)=(gh)A. If A ⊆ B and g ∈ G, then gA ⊆ gB. The proofs of the lemma above will be left as exercise. We often use easy results like these without mentioning explicitly. 5.2. Lemma. Let φ : G → H be a group homomorphism and let N = ker(φ). Then φ−1(φ(g)) = gN = Ng for all g ∈ G. Proof. Take g ∈ G. Let x ∈ φ−1(φ(g)) or in other words, φ(x)= φ(g). Then −1 −1 −1 φ(g x)= φ(g) φ(x)= φ(g) φ(g) = eH, so g−1x ∈ N, or in other words x ∈ gN. Thus φ−1(φ(g)) ⊆ gN. Conversely, if x ∈ gN, −1 then x = gn for some n ∈ N, so φ(x) = φ(g)φ(n) = φ(g)eH = φ(g), so x ∈ φ (φ(g)). Thus gN ⊆ φ−1(φ(g)). This proves φ−1(φ(g)) = gN. A similar argument shows that φ−1(φ(g)) = Ng (Exercise: Write out this argument). 5.3. Definition. Let G be a group. A subgroup N of G is called a normal subgroup if gN = Ng for all g ∈ G. In this case we need not distinguish between left and right cosets. So we simply talk of cosets of N. 5.4. Lemma. Let G be a group and N be a subgroup of G. The following are equivalent. (a) gN = Ng for all g ∈ G. (b) gNg−1 = N for all g ∈ G. (c) If g ∈ G and n ∈ N, then gng−1 ∈ N. Proof. Suppose (a) holds. Then gNg−1 = (gN)g−1 = (Ng)g−1 = N(gg−1) = N. Thus (a) implies (b). Clearly (b) implies (c). Assume (c). Let x ∈ gN. Then x ∈ gn for some n ∈ N. So x = gng−1g ∈ Ng since gng−1 ∈ N by our assumption. Thus gN ⊆ Ng. Similarly show that Ng ⊆ gN. So gN = Ng. Thus (c) implies (a). 5.5 (Motivating the construction of a quotient). Lemma 5.2 says that kernels of homomorphisms are normal subgroups. Conversely, we shall see that all normal subgroups appear as kernels of homomorphisms. Let G be a group and let N be a normal subgroup of G. We shall construct a onto group homomorphism π : G → Q such that ker(π)= N. This group Q will be called the quotient of G by the normal subgroup N and will be denoted by Q = G/N. To motivate the construction, suppose we have an onto group homomorphism π : G → Q and let ker(π) = N. Let q ∈ Q. Since π is onto, we can pick an element g ∈ G such that π(g) = q. Then Then lemma 5.2 gives us π−1(q) = gN. In words, this says that the preimages of elements of Q are just the cosets of N. So the elements of Q are in one to one correspondence with the set of cosets of N in G. Also verify that if q1 and q2 are two elements −1 −1 −1 of Q such that π (q1)= g1N and π (q2)= g2N, then π (q1q2)= g1g2N (Exercise: verify this). This suggests the definition that follows. 5.6. Definition. Let G be a group and N be a normal subgroup. Let G/N denote the set of cosets of N in G. Let C and D be two cosets of N in G. We say that an element c ∈ G 8 is a representative of the coset C if c ∈ C. Choose a representatitve c of the coset C and a representative d of the coset D.

Claim: The coset cdN only depends on the cosets C and D and does not depend on the choice of the representatives c and d.

Proof. Let c1 be another representative of C and d1 be another representative of D. Then −1 −1 −1 −1 c1c ∈ N and d1d ∈ N. Since N is normal, c(d1d )c ∈ N as well. So −1 −1 −1 −1 −1 −1 (c1d1)(cd) = c1d1d c =(c1c )c(d1d )c ∈ N.

So c1d1 ∈ Ncd = cdN. So c1d1N = cdN. Define a binary operation on G/N as follows: Given two cosets C and D, we choose coset representatives c and d of C and D respectively and let let (C.D) to be the coset cdN. The claim we just proved imply that this procedure gives a well defined binary operation on G/N. Let us reiterate the definition of the binary operation in slightly different words. If c ∈ C and d ∈ D, then C = cN and D = dN. So our definition says (cN).(dN)= cdN. Let us also reiterate that we have already argued this is a well defined operation on G/N and this argument required the fact that N is a normal subgroup. Next one verifies that G/N with the binary operation just defined becomes a group where the identity is the coset N and inverse of gN is g−1N. This group G/N is called the quotient of G by N. There is a canonical onto function from π : G → G/N given by π(g) = gN. Verify that π is a homomorphism and ker(π)= N.

5.7. Example. Verify that Zn = Z/nZ.

6. Constructing homomorphisms 6.1. Lemma. Let G be a group and a ∈ G. Then there exists a homomorphism φ : Z → G given by φ(n) = an for all n ∈ Z. This φ is the unique homomorphism from Z to G such that φ(1) = a. sketch of proof. Deﬁne φ(n) = an for all n ∈ Z. Verify that φ : Z → G is a homomorphism such that φ(1) = a. Let φ′ : Z → G be any homomorphisms such that φ′(1) = a. Then φ′(n)= an = φ(n) for all n ∈ Z. So φ′ = φ. The above lemma says that constructing homomorphisms from Z to any group is easy. You can send 1 to anything you want and the homomorphism is completely determined once you decide where 1 goes.

6.2. Corollary. The map φ(k) = k mod n is the unique homomorphism φ : Z → Zn such that φ(1) = 1 mod n. The kernel of this homomorphism is nZ.

6.3. Lemma. Let φ1 : G → H1 and φ2 : G → H2 be two homomorphisms. Then there is a homomorphism φ : G → H1 × H2 given by φ(G)=(φ1(G),φ2(G)). 6.4. Theorem. Let A,B,C be three groups and let f : A → B and g : A → C be two homomorphisms. (i) if there exists a homomorphism h : B → C such that h ◦ f = g, then ker(f) ⊂ ker(g). 9 (ii) Suppose f is onto and ker(f) ⊂ ker(g). (a) Then there exists a unique homomorphism h : B → C such that h ◦ f = g. (b) If g is onto, then so is h. (c) If ker(g) = ker(f), then h is one to one. (d) If g is onto and ker(g) = ker(f), then h is an isomorphism. Proof. (i) Exercise. (ii) (a) Uniqueness: Suppose h, h′ be two homomorphisms such that h ◦ f = g = h′ ◦ f. Pick b ∈ B. Since f is onto there exists a ∈ A such that f(a) = b. then h(b) = h(f(a)) = g(a) = h′(f(a)) = h′(b). So h = h′. Thus there can be atmost one h with the given properties. Existence: Let b ∈ B. Suppose a, a′ ∈ A such that f(a)= f(a′)= b. Then f(a−1a′)= e, so (a−1a′) ∈ ker(f), so (a−1a′) ∈ ker(g), so g(a−1a′) = e, so g(a) = g(a′). Thus we observe that “g takes the same value on all the elements of A that map to b”. Given b ∈ B, we can pick a ∈ A such that f(a) = b (since f is onto) and then deﬁne h(b) = g(a). Because of our observation, this procedure gives us a well deﬁned function h : B → C. Verify that h is a homomorphism. Now suppose a ∈ A. Let b = f(a). So a is an element of A such that f(a)= b. So h(b)= g(a), or h(f(a)) = g(a). So h ◦ f = g. This proves part (a). (b) Now suppose g is onto. Let c ∈ C. Then there exists a ∈ A such that g(a) = c. It follows that h(f(a)) = c, so h is onto. This proves part (b). (c) Now suppose ker(f) = ker(g). To show h is one to one, it is enough to verify that ker(h) = eB. Let b ∈ ker(h), that is, h(b) = eC . Pick a ∈ A such that f(a) = b. Then g(a)= h(f(a)) = h(b)= eC , so a ∈ ker(g) = ker(f), so b = f(a)= eB. So ker(h) = eB. This proves part (c). Part (d) follows from part (b) and (c). 6.5. Corollary. Let m, n be positive integers. Then there exists a unique homomorphism h : Zmn → Zm such that h(1 mod mn) = 1 mod m. sketch of proof. Let f : Z → Zmn and g : Z → Zm be the homomorphisms f(k) = k mod mn and g(k) = k mod m. Then ker(f) = mnZ ⊆ mZ = ker(g). Now apply the above theorem. It is convenient to represent the statement of theorem 6.4 in a diagrammatic language. Let us denote an onto homomorphism f : A → B by a double headed arrow f : A ։ B. The theorem 6.4 says that g / A ❅ C ❅❅ ❅❅ ❅❅ f ❅ B with ker(f) ⊆ ker(g), there exists a unique h : B → C such that the following diagram commutes: g / A ❅ ? C (3) ❅❅ ⑦? ❅❅ ⑦ ❅❅ ⑦ f ❅ ⑦ h B Saying that the diagram in 3 commutes is a pictorial way of saying that g = h ◦ f.

10 6.6. Theorem (The ﬁrst isomorphism theorem). Let ψ : G → H be a group homomorphism. Let K = ker(ψ). Let φ : G → G/K be the natural quotient homomorphism: φ(g) = gK. Then there is a unique isomorphism η : G/K → ψ(G) such that ψ = η ◦ φ. Proof. Recall that kernels of homomorphisms are normal subgropus, so K is normal in G. From given data, we get the following diagram of homomorphisms

ψ G / / ψ(G) ❈❈ ❈❈ ❈❈ ❈❈ φ ❈! ! G/K Note that the horizontal arrow is simply the given map ψ, but we consider it as a map to the subgroup ψ(G) of H. So ψ : G → ψ(G) is an onto homomorphism. From the deﬁnition of the quotient map, we have ker(φ)= K = ker(ψ). So by theorem 6.4(d), we get an isomorphism η : G/K → ψ(G), such that the following diagram commutes:

ψ G / / ψ(G) ❈❈ ; ❈❈ ✇✇; ❈❈ ≃ ✇✇ ❈❈ ✇✇η φ ❈! ! ✇✇ G/K Saying that the above diagram commutes is just a pictorial way of saying ψ = η ◦ φ. Let G be a group and N be a normal subgroup. The quotient homomorpshim G → G/N collapses the normal subgroup N to the identity in G/N. In this way, considering the quotient group G/N lets us forget the structure inside N. The correspondence theorem below says that taking quotient by N “retains the subgroup structure in G above N”. So passing to the quotient group G/N lets us disregard what was happening inside N and focus solely on what happens “above” N. Thus, the quotient construction becomes an important simplifying tool. 6.7. Theorem (the correspondence theorem). Let N be a normal subgroup of G. Then there is a natural inclusion preserving bijection between {set of subgroups of G containing N }→{set of subgroups of G/N} Under this bijection, a subgroup H of G such that H ⊃ N corresponds to the subgroup H/N of G/N. Under the above correspondence, normal subgroups on the left hand side correspond to normal subgroups on the right hand side. The proof of the correspondence theorem is left as a routine exercise. We want to move on to the second and the third isomorphism theorems. These theorems let us compare the subgroup structures in groups with subgroup structures in various quotients. Before stating them we need a lemma. The proof of the lemma is left as an exercise. 6.8. Lemma. Let H and K be two subgroups of a group G. Then HK is a subgroup of G if and only if HK = KH. Proof. Suppose HK = KH. Let c,c′ ∈ HK. Then there exists h, h′ ∈ H and k,k′ ∈ K such that c = hk and c′ = h′k′. Note that kh′ ∈ KH = HK, so kh′ = h′′k′′ for ome h′′ ∈ H and 11 k′′ ∈ K. Then cc′ = hkh′k′ = hh′′k′′k′ ∈ HK since hh′′ ∈ H and k′′k ∈ K. Thus HK is closed under multiplication. Let c ∈ HK. Then there exists h ∈ H and k ∈ K such that c = hk. Now c−1 = k−1h−1 ∈ KH = HK. So HK is closed under taking inverses. Clearly eG belongs to HK. It follows that HK is a subgroup. This proves one implication. The other implication is left as an exercise. 6.9. Theorem (The second isomorphism theorem). Let N be a normal subgroup of G. Let H be a subgroup of G. Then HN is a subgroup of G and H ∩ N is a normal subgroup of H and H/H ∩ N ≃ HN/N. sketch of proof. Since N is normal in G, we hav gN = Ng for all g ∈ G. So HN = NH. By lemma 6.8, it follows that HN is a subgroup of G. Verifying H ∩ N is a normal subgroup of H is left as a routine exercise. Now let ψ be the composition of homomorphisms: quotient map H −−−−−−−−→inclusion HN −−−−−−−−−−→ HN/N. So ψ : H → HN/N is given by ψ(h) = hN. Note that ψ is a homomorphism since both the inclusion and the quotient map are homomorphisms. Now verify that ψ is onto and ker(ψ)= H ∩ N. Then the second isomorphism theorem follows from the ﬁrst isomorphism theorem. 6.10. Theorem (The third isomorphism theorem). Let H and N be normal subgroups of G G/N G with N ⊂ H ⊂ G. Then there is a natural isomorphism H/N ≃ H . Proof. Consider the following diagram of homomorphisms:

φH G / / G/H ❇❇ ❇❇ ❇❇ ❇❇ φN ❇! ! G/N where both the arrows φN and φH are the natural quotient homomorphisms. Note that ker(φN ) = N ⊂ H = ker(φH ). So by theorem 6.4(c), we have a onto homomorphism ψ : G/N → G/H such that the following diagram commutes:

φH G / / G/H ❇❇ ; ❇❇ ✇✇; ; ❇❇ ✇✇ ❇❇ ✇✇ φN ❇! ! ✇✇ ψ G/N

If g ∈ G, then one has ψ(gN)= ψ ◦ φN (g)= φH (g)= gH. Now verify that ker(ψ)= H/N and apply the ﬁrst isomorphism theorem.

Let G1 and G2 be two groups. Let us consider the product group G1 × G2. The product comes with natural projection maps π1 : G1 × G2 → G1 and π2 : G1 × G2 → G2 given by π1(g1,g2) = g1 and π2(g1,g2) = g2. It is easy to verify that π1 and π2 are onto homomorphisms. The next theorem tells us how to construct homomorphism to a product.

6.11. Theorem. Let G1,G2,G be groups. Given homomorpshims f1 : G → G1 and f2 : G → G2, there exists a unique homomorphism f : G → G1 × G2 such that π1 ◦ f = f1 and π2 ◦ f = f2 (where π1 and π2 are the natural projections. 12 sketch of proof. Deﬁne f : G → G1 × G2 by f(g)=(f1(g), f2(g)) and verify that f is a homomorphism and that π1 ◦ f = f1 and π2 ◦ f = f2 6.12. Theorem. Let H,K be two normal subgroups of a group G such that H ∩ K = {e}. Then show that G is isomorphic to a subgroup of G/H × G/K. sketch of proof. Let φH : G → G/H and φK : G → G/K be the natural quotient homomorphisms. By 6.11, we get a homomorphism φ : G → G/H × G/K, given by

φ(g)=(φH (g),φK(g))=(gH,gK). Let g ∈ Ker(φ). Then (gH,gK) is the identity in G/H × G/K. This means gH is the identity in G/H and gK is the identity in G/K. To say gH is the identity in G/H, means gH = H, so g ∈ H. Similarly, we get g ∈ K. So g ∈ H ∩ K = {e}. So ker(φ) = {e} and hence φ is a one to one homomorpshim. So φ : G → G/H × G/K gives an isomorphism φ : G → φ(G) and φ(G) is a subgroup of G/H × G/K. 6.13. Theorem. Let H and K be two subgroups of a ﬁnite group G such that K is normal in G, H ∩ K = {e} and HK = G. Then show that |G| = |H|·|K|. Proof. By the second isomorphism theorem, we know that HK is a subgroup of G and H ∩K is a normal subgroup of H and that H/H ∩ K ≃ HK/K. Now since H ∩ K = {e} and HK = G, we get H ≃ H/{e} = H/H ∩ K ≃ HK/K ≃ G/K. So |H| = |G/K| = |G|/|K|. Infact, given the hypothesis of 6.13, it follows that G ≃ H × K (Exercise: show this). In this situation, one says that G is the internal direct product of H and K. 6.14. Theorem. Let H,K be two normal subgroups of a ﬁnite group G such that H∩K = {e} and HK = G. Then show that G is isomorphic to G/H × G/K. Proof. By 6.12, we have an one to one homomorphism φ : G → G/H × G/K given by φ(g)=(gH,gK). By 6.13 we have |G| = |K|·|H|. So |G/H| = |G|/|H| = |K| and |G/K| = |G|/|K| = |H|. So |G/H × G/K| = |G/H|·|G/K| = |K|·|H| = |G|. By 6.12, we have an one to one homomorphism φ : G → G/H × G/K given by φ(g) = (gH,gK). Since G and G/H × G/K have the same size, the map φ must be onto as well. So φ : G → G/H × G/K is an isomorphism. The following result, known as the Chinese remainder theorem is a special case of the above considerations, but we shall state and prove it directly. 6.15. Theorem. Let m, n be two nonzero integers. If m and n are relatively prime, then Zmn ≃ Zm × Zn.

Proof. By 6.5, we have a natural homomorphisms Zmn → Zm and Zmn → Zn given by taking (k mod mn) to (k mod m) and to (k mod n) respectively. By 6.11, this gives a homomorphism f : Zmn → Zm × Zn given by f(k mod mn)=(k mod m, k mod n) 13 Let k ∈ Z such that (k mod mn) ∈ ker(f), then (k mod m, k mod n) = (0 mod m, 0 mod n). In other words m and n divide k. Since m and n are relatively prime, this means mn divides k, so k mod mn is the identity of Zmn. So ker(f) is just the identity of Zmn, hence f : Zmn → Zm × Zn is one to one. Since |Zmn| = mn = |Zm|·|Zn| = |Zm × Zn|, the map f must be onto as well. So f is an isomorphism. 6.16. Corollary. Let m and n be relatively prime non-zero integers. Then (a) U(mn) ≃ U(m) × U(n). (b) For a positive integer n let φ(n) be the Euler Totient function deﬁned as the number of integers k such that 1 ≤ k < n and k is relatively prime to n. For example φ(p)=(p − 1) if p is a prime number. One has φ(mn)= φ(m)φ(n). e1 er (c) Let n be a positive integer. Let n = p1 ··· pr be the prime factorization of n. Then −1 −1 φ(n)= n(1 − p1 ) ··· (1 − pr ).

sketch of proof. Consider the isomorphism f : Zmn → Zm × Zn given in 6.15. Since f is onto, given any (α, β) ∈ U(m) × U(n), there exists k ∈ Z such that f(k mod mn)=(α, β). This means (k mod m)= α and (k mod n)= β are relatiely prime to m and n respectively, so k is relatively prime to mn, that is, (k mod mn) ∈ U(mn). Conversely, if (k mod mn) ∈ U(mn), then f(k mod mn)=(k mod m, k mod n) ∈ U(m) × U(n). So the restriction of f to U(mn) gives an onto map f : U(mn) → U(m) × U(n). Since f is one to one, so is the restriction. Finally verify that f is also a homomorphism for the multiplicative groups U(mn) to U(m) × U(n). This proves part (a). Part (b) follows immediately, since |U(n)| = φ(n). Part (c) follows from part (b) and the easy observation that φ(pr)= pr(1 − p−1) for a prime number p.

14 7. Few more exercises 7.1. Exercise. Let p be a prime. Let G be a group of order p2. Show that G ≃ Z/p2Z or G ≃ Z/pZ × Z/pZ. Proof. Let G∗ denote the set of non-identity elements of G. If G has an element of order p2, then we G ≃ Z/p2Z. Otherwise each element of G∗ must have order p. So all nontrivial proper subgroups of G are isomorphic to Z/pZ and any two of these distinct subgroups intersect trivially. Pick x ∈ G∗. Write H = hxi and pick y ∈ G − H. We shall argue that either hxi or hyi is normal in G. Suppose hxi is not normal in G. Then we claim that H has exactly p distinct conjugates subgroups {yrHy−r : 0 ≤ r

7.2. Exercise. Describe the homomorphisms from Z24 to Z18. sketch of proof. Let π : Z → Z24 be the natural quotient homomorphism: π(n)= n mod 24. For k =0, 1, ··· , 5, deﬁne the homomorphisms gk : Z → Z18 by gk(r)=3rk mod 18. Notice that gk(24m) = 72mk mod18 = 0mod18. So ker(π) = 24Z ⊆ ker(gk). Thus there exists fk : Z24 → Z18 such that the following diagram commutes:

gk Z / Z ❆ = 18 ❆❆ ③③= ❆❆ ③③ π ❆❆ ③③ ❆ ③③ fk Z24

In other words we have found six homomorphisms fk : Z24 → Z18 given by fk(m mod 24) = fk(π(m)) = gk(m)=3km mod 18 for 0 ≤ k ≤ 5. Now suppose f : Z24 → Z18 be any homomorphism. Let f(1 mod 24) = m mod 18. Then Note that 0 mod 18 = f(24 mod 24) = 24m mod 18, so 24m must be a multiple of 18, so m must be a multiple of 3. Hence f(1 mod 24) = 3k mod 24 for some k. So for any r ∈ Z, we have f(r mod 24) = rf(1 mod 24) = 3rk mod 18 = fk(r mod 24). So f = fk for some k. It follows that fk’s are all the possible homomorphisms from Z24 to Z18. 15 7.3. Exercise. (a) Let τ1 and τ2 be two transpositions in Sn. Show that τ1τ2 can be written as a product of 3-cycles. (b) Prove that every element of An can be written as product of 3-cycles.

Proof. (a) Let τ1 =(i j) and τ2 =(k l). Suppose {i, j} and {k,l} are disjoint. Then notice that τ1τ2 = (ijk)(jkl); so τ1τ2 is a product of 2-cycles. Next suppose {i, j} and {k,l} have one element in common. By renaming the elements if necessary, without loss, we may assume that the common element is j = k. Then τ1τ2 = (j l i) is a three cycle. The remaining possibility is τ1 = τ2. In this case τ1τ2 is the identity permutation, which is also the (empty) product of 3-cycles. This proves part (a). (b) Let σ be any permutation in An. Since σ is an even permutation, we can write σ = τ1τ2 ··· τ2n where τj are transpositions. From part (a) we know that for j = 1, ··· , n, the product τ2j−1τ2j can be written as a product of 3-cycles. Multiplying these together, we ﬁnd that σ can also be written as a product of 3-cycles.

3 3 7.4. Exercise. Consider the group R . Let H = {(x1, x2, x3) ∈ R : x1 +2x2 − x3 = 0}. Show that H is a normal subgroup of R3. Show that R3/H ≃ R. Proof. The identity of the additive group R3 is 0 = (0, 0, 0). Notice that 0 ∈ H so H =6 ∅. Let x =(x1, x2, x3) and y =(y1,y2,y3) be two elements of H. Then x1 + 2x2 − x3 = 0 and y1 + 2y2 − y3 = 0. It follows that the coordinates of z = x − y =(x1 − y1, x2 − y2, x3 − y3) satisfy

(x1 − y1)+2(x2 − y2) − (x3 − y3)=(x1 +2x2 − x3)+(y1 +2y2 − y3)=0. So x − y ∈ H if x, y ∈ H. This directly proves that H is a subgroup of R3. Since R3 is abelian, any subgroup is automatically normal. Alternatively, we can argue as follows: Now deﬁne f : R3 → R by

f(x1, x2, x3)= x1 +2x2 − x3.

3 Let x =(x1, x2, x3) and y =(y1,y2,y3) be two elements of R . Then we verify that

f(x + y)= f(x1 + y1, x2 + y2, x3 + y3)

=(x1 + y1)+2(x2 + y2) − (x3 + y3)

=(x1 +2x2 − x3)+(y1 +2y2 − y3) = f(x)+ f(y). So f is a group homomorphism. Looking at the definition of H, we notice H = ker(f). Since the kernel of any homomorphism is a normal subgroup, we find that H is a normal subgroup of R3. Given any x ∈ R, we notice that f(x, 0, 0) = x, so f is an onto homomorphism. Thus by the first isomorphism theorem, we get an isomorphism R3/H ≃ R. 7.5. Exercise. Let G be a group and a, b ∈ G. Suppose a has order m and b has order n. If gcd(m, n)=1, then show that hai∩hbi = {e}. Proof. Let x ∈hai∩hbi. Since x ∈hai, we have x = at for some t ∈ Z. Since am = e, it follows that xm = atm = e. Similarly since x ∈hbi, we get xn = e. Since gcd(m, n) = 1, there exists r, s ∈ Z such that mr + ns = 1. It follows that x = xmr+ns =(xm)r(xn)s = eres = e. 16 7.6. Exercise. Let G be a group. An element of G is called a commutator if it has the form aba−1b−1 for some a, b ∈ G. (a) If c is a commutator and g ∈ G. Then show that c−1 is a commutator and that gcg−1 is a commutator. (b) Let G′ be the set of all elements of G that can be written as a product of commutators. Then show that G′ is a normal subgroup of G. (c) Let N be a normal subgroup of G. Show that G/N is abelian if and only if N contains G′. Proof. (a) Let c be a commutator. Then c has the form c = aba−1b−1 for some a, b ∈ G. Note that c−1 = bab−1a−1. So c−1 is also a commutator. Also, notice that gcg−1 = gaba−1b−1g−1 =(gag−1)(gbg−1)(gag−1)−1(gbg−1)−1. Thus gcg−1 is also a commutator. This proves part (a). (b) Note that e = eee−1e−1 can be written as a commutator. Let x ∈ G′. Then x = −1 −1 −1 c1c2 ··· cn where each cj is a commutator. Then x = cn ··· c1 . From part (a), we know −1 −1 that each cj is a commutator. It follows that x is also a product of commutators, so −1 ′ ′ x ∈ G . Now let x, y ∈ G . Then we can write x = c1 ··· cn and y = d1 ··· dm where each cj and dj is a commutator. Then xy = c1 ··· cnd1 ··· dm is also a product of commutators, so xy ∈ G′. Thus G′ is a subgroup of G. Now suppose x ∈ G′ and g ∈ G. Then we can write −1 −1 −1 x = c1 ··· cn where each cj is a commutator. Then gxg =(gc1g ) ··· (gcng ). From part −1 −1 (a) we know that each gcjg is a commutator. So we find that gxg is also a product of commutators, so gxg−1 ∈ G′. Thus G′ is a normal subgroup of G. (c) Let N be a normal subgroup of G such that G/N is abelian. Let a, b ∈ G. Then in the factor group G/N, we have abN = aN.bN = bN.aN = baN. It follows that (ab)(ba)−1 ∈ N, that is, aba−1b−1 ∈ N. So N contains all the commutators. Since N is a subgroup, it follows that N contains all the products of commutators, so N contains G′.