Solutions to Assignment 1

1. Determine all group structures on a set of n elements for 2 ≤ n ≤ 6. Determine the subgroups, quotient groups of these groups and also their Jordan-Holder series. Solution:

n = 1: The trivial group.

n = 2: The group Z2.

n = 3: The group Z3.

n = 4: Groups Z4 and Z2 × Z2. Jordan Holder series are given by, Z4 ⊃ Z2 ⊃ {e},Z2 × Z2 ⊃ Z2 ⊃ {e}.

n = 5: The group Z5.

n = 6: The groups Z6,S3. Jordan Holder series, Z6 ⊃ Z3 ⊃ {e}, S3 ⊃ H ⊃ {e}, where H is the subgroup of order 3 containing the identity and the 3-cycles.

2. Let A and B be subgroups of a group G.

(a) Prove that for AB to be a group it is necessary and suﬃcient that A and B are permutable, that is AB = BA. Solution: Suppose that AB is a group. Let a ∈ A, b ∈ B. Then, ab ∈ AB. −1 −1 −1 −1 −1 But, (ab) = b a must also be in AB. Hence, b a = a1b1 for some −1 −1 a1 ∈ A, b1 ∈ B. Inverting again, ab = b1 a1 ∈ BA. Hence, AB ⊂ BA and by symmetry BA ⊂ AB.

Now suppose, AB = BA. Let a1, a2 ∈ A and b1, b2 ∈ B. Now consider, a1b1a2b2. Since, BA = AB, b1a2 = a3b3 for some a3 ∈ A and b3 ∈ B. Thus, AB is closed −1 −1 −1 under multiplication. Also, (a1b1) = b1 a1 ∈ BA = AB. Thus, AB is also closed under taking inverses. (b) Prove that if A and B are permutable and C is a subgroup containing A, A is permutable with B ∩ C and A(B ∩ C) = C ∩ (AB).

Solution: Let a ∈ A and b ∈ B ∩ C. Since, A, B are permutable, ab = b1a1 for some b1 ∈ B and a1 ∈ A. But since, A ⊂ C, a, b, ab, a1 are all in C, which implies that b1 ∈ C. Hence, A is permutable with B ∩ C. Clearly, A(B ∩ C) ⊂ C ∩ (AB). Now, let c = ab ∈ C ∩ (AB) for some a ∈ A, b ∈ B, c ∈ C. Now, b = a−1c ∈ C because a ∈ C. Hence, ab ∈ A(B ∩ C).

3. If a subgroup of a group G has index 2, show that it is a normal subgroup of G. Solution: Let H be a subgroup of G of index 2 and let g ∈ G \ H. Since, there are only two cosets and gH 6= H, gH = G \ H. But the same is true for Hg. Hence, gH = Hg for every g, proving that H is normal.

4. Let G be a group such that all elements of G other than the identity has order 2. Show that:

1 Solutions to Assignment 1

(a) G is commutative. Solution: Let a, b ∈ G. Then, (ab)2 = abab = e. Hence, ab = b−1a−1. But, b2 = e ⇒ b = b−1 and similarly a = a−1. Hence, ab = ba. (b) If G is finite, order of G is a power of 2. Solution: From the book, for a finite Abelian group the order of G must divide some power of any exponent of G. Clearly, 2 is an exponent and hence order of G must be a power of 2. 5. Let G be a group such that, for a fixed integer n > 1, (xy)n = xnyn for all x, y ∈ G. (n) n n Let G = {x |x ∈ G}, and G(n) = {x ∈ G|x = e}.

(n) (a) Prove that G ,G(n) are normal subgroups of G. n Solution: It is easy to see by induction that for any g1, . . . , gk ∈ G,(g1 . . . gk) = n n n n n (n) g1 . . . gk . It is also clear from the property that (xy) = x y , that G ,G(n) are subgroups. Now, let g, x ∈ G and consider the element gxng−1. Clearly,

n −1 −1 −1 −1 n (n) gx g = (gxg1)(gxg ) ... (gxg ) = (gxg ) ∈ G . This shows that, G(n) is a normal subgroup. n −1 n n n −n Now let y ∈ G(n). Hence, y = e. Now, (gyg ) = (g y g ) = e. Hence, −1 gyg ∈ G(n), which show that G(n) is a normal subgroup. (n) (b) If G is ﬁnite then the order of G is equal to the index of G(n). Solution: We will establish a bijection between the elements of G(n) and the left n (n) cosets of G(n). Namely, let x = g ∈ G map to gG(n). Clearly, the map is onto. Now, suppose that g1G(n) = g2G(n). Then, g1y1 = g2y2 for some y1, y2 ∈ G(n). n n n n n n n n n n But, y1 = y2 = e. Hence, (g1y1) = g1 y1 = g1 = (g2y2) = g2 . Hence, g1 = g2 proving that the map is 1-1. (c) Show that for all x, y ∈ G, x1−ny1−n = (xy)1−n and deduce that xn−1yn = ynxn−1. Solution: x1−ny1−n = xx−ny−ny = xx−1y−1ny = xx−1(y−1x−1)n−1y−1y = (y−1x−1)n−1 = (xy)−1n − 1 = (xy)1−n. Also, xn−1yn = (x−1)1−n(y−1)1 − ny = (x−1y−1)1−ny = (yx)n−1y = (yx)n−1yxx−1 = (yx)nx−1 = ynxn−1. (d) Conclude from the above that the set of elements of G of the form xn(n−1) generates a commutative subgroup of G. Solution: Let x, y ∈ G. Now,

xn(n−1)yn(n−1) = (xn−1)n(yn−1)n = (xn−1yn−1)n =

2 Solutions to Assignment 1

(xn−1yny−1)n = (ynxn−1y−1)n = yn2 x(n−1)ny−n = yn2 (xn)n−1(y−1)n = yn2 (y−1)n(xn)n−1 = yn(n−1)xn(n−1). Thus, he set of elements of G of the form xn(n−1) commute with each other and hence, they generate a commutative subgroup.

6. Let G be a group. If S is a simple group, S is said to occur in G is there exist 0 0 0 ∼ two subgroups H,H of G, with H C H such that H /H = S. Let in(G) be the isomorphism classes of simple groups occuring in G.

(a) Prove that, in(G) = ∅ ⇔ G = {e}. Solution: Clearly if G = {e} then in(G) = ∅. Now, suppose that in(G) = ∅. If G 6= {e} then there exists an element g ∈ G, g 6= e. Consider the cyclic subgroup 2 ∼ hgi generated by g. Now, if hgi is inﬁnite then hgi/hg i = Z2, which must belong to in(G). If hgi is ﬁnite, then it is simple if |hgi| is prime. Otherwise, let H be a maximal proper subgroup of hgi, and hgi/H must be simple and belong to in(G). (b) If H is a subgroup of G, show that in(H) ⊂ in(G); if moreover, H is normal, then

in(G) = in(H) ∪ in(G/H).

Solution: If H is normal it follows easily from the third isomorphism theorem, that in(H) ∪ in(G/H) ⊂ in(G).

Now, let S be a simple group, such that there exist two subgroups H1,H2 of G, ∼ with H1 C H2, and H2/H1 = S. Now, H2 ∩ H C H2, and hence H1(H2 ∩ H) C H2. Since, H2/H1 is simple, there is no proper normal subgroup of H2 properly containing H1. ∼ Hence, either H1(H2 ∩ H) = H2 or H1(H2 ∩ H) = H1. In the ﬁrst case, S = ∼ ∼ H2/H1 = H1(H2 ∩ H)/H1 = (H2 ∩ H)/(H1 ∩ H2 ∩ H) ∈ in(H), where we used the second isomorphism theorem. 0 ∼ 0 In the second case, H = H2 ∩H ⊂ H1 and H2H/H = H2/H contains a subgroup isomorphic to S. Hence, S ∈ in(G/H).

ii. Every subgroup of G1 ×G2 is of the form H1 ×H2 with H1 ⊂ G1 and H2 ⊂ G2.

Solution: Let in(G1) ∩ in(G2) = ∅. Let H be a subgroup of G1 × G2 and H1 = p1(H) and H2 = p2(H2) where p1, p2 are the projection homomorphisms. ∼ We claim that, H = H1 × H2. 0 −1 0 −1 0 Let H1 = p1(p2 (e2) ∩ H) and H2 = p2(p1 (e1) ∩ H). It is easy to verify that, H1 0 is a normal subgroup of H1 and similarly, H2 is a normal subgroup of H2.

3 Solutions to Assignment 1

Consider, the homomorphism

0 0 φ1 : H1 → H/(H1 × H2)

−1 defined by, φ1(h1) = (h1, h2), where (h1, h2) is any element of p1 (h1)∩H. This is 0 −1 0 −1 well-defined since for any other element (h1, h2) of p1 (h1)∩H,(h1, h2)(h1, h2) = 0−1 0 0 0 0 (e1, h2h2 ) ∈ H1 × H2. Also, h1 ∈ ker(φ) iff h1 ∈ H1. Hence, ker(φ) = H1. 0 0 Moreover, let (h1, h2) ∈ H/(H1 × H2). Clearly, φ1(h1) = (h1, h2) and hence φ1 is 0 ∼ 0 0 surjective. Hence, H1/H1 = H/(H1 × H2). By symmetry, 0 ∼ 0 0 ∼ 0 H1/H1 = H/(H1 × H2) = H2/H2.

0 0 Now, in(H1/H1) ⊂ in(H1) ⊂ in(G1) and in(H2/H2) ⊂ in(H2) ⊂ in(G2). Since, 0 ∼ 0 0 0 in(G1)∩in(G2) = ∅, and H1/H1 = H2/H2, it follows that in(H1/H1) = in(H2/H2) = 0 0 ∅, and using part (a) this implies that H1 = H1 and H2 = H2. It is easy now to deduce that H = H1 × H2.

In order to prove the converse, suppose there exists S ∈ in(G1) ∩ in(G2). This 0 0 implies that there exists subgroups H1 C H1 of G1 and H2 C H2 of G2, such that ∼ 0 ∼ 0 0 0 S = H1/H1 = H2/H2. Choose an isomorphism φ : H1/H1 → H2/H2. ¯ ¯ Let H = {(h1, h2)|h1 ∈ H1, h2 ∈ H2, h2 = φ(h1)}.

Then, it is easy to check that H is a subgroup of H1 × H2, p1(H) = H1 and ¯ ¯ p2(H) = H2. However for any h1 ∈ H1, there exists h2 ∈ H2 such that h2 6= φ(h1) 0 since, H1/H1 is not trivial. Hence, H is not of the form P1 × P2 for subgroups P1 of G1 and P2 of G2. 7. Let n be a positive integer, and consider the set G of nonzero integers less than or equal to n, which are relatively prime to n. The number of elements of G is called the Euler phi-function, denoted ϕ(n). For example, ϕ(1) = 1, ϕ(2) = 1, ϕ(3) = 2, ϕ(4) = 2, etc.

(a) Show that G is a group under multiplication mod n. Solution: Let x, y ∈ G. Since they are relatively prime to n, so is their product. Consequently xy ≡ z mod n for some z ∈ G. The element 1 serves as the identity and since G is ﬁnite, we may use Problem 1 above. Suppose x, y, a ∈ G with xa ≡ ya mod n. Then n divides xa − ya = (x − y)a. Since a is relatively prime to n, we must have n|(x − y). But then x and y are both positive integers less than or equal to n, so they must be equal. (b) If m and n are relatively prime positive integers, show that

mϕ(n) ≡ 1 mod n.

Solution: Since m is relatively prime to n, there exists x ∈ G with x ≡ m mod n. The order of x divides |G| = ϕ(n), and so xϕ(n) ≡ 1 mod n. This implies mϕ(n) ≡ 1 mod n.

4 Solutions to Assignment 1

8. Let n be a positive integer. Show that X n = ϕ(d), d|n where the sum is taken over all positive integers d which divide n. (Hint: Recall that a cyclic group of order n has a unique subgroup of order d for each integer d which divides n.) Solution: Let G = Z/hni. Note that if x ∈ G, then the order of x divides n. X n = |G| = number of elements of G which have order d d|n X = number of generators for the unique subgroup of G of order d d|n X = ϕ(d). d|n

9. Let G be a ﬁnite group with identity e, which has the property that for every positive integer d, the number of elements x of G with xd = e is at most d. Show that G must be cyclic. Solution: Let |G| = n, and a ∈ G be an element of order d. Then the cyclic group hai = {a, a2, . . . , ad} has d distinct elements satisfying xd = e, and so these must be the only elements x with xd = e. Consequently if G has at least one element of order d, then it has precisely ϕ(d) elements of order d. But X X n = number of elements of G which have order d ≤ ϕ(d) = n, d|n d|n and so for each d dividing n, the number of elements of G which have order d must be precisely ϕ(d). In particular, G has ϕ(n) elements of order n, and therefore is a cyclic group. 10. Recall from class the deﬁnition of a nilpotent group. Let f be a homomorphism from a nilpotent group G to a group G0. Prove that if the restriction of f to the center of G is injective, then f is injective. Solution: Since G is nilpoten there exists a sequence,

G = G0 ⊃ G1 ⊃ · · · ⊃ Gn = {e}

such that for each i, 1 ≤ i ≤ n − 1, [G, Gi] ⊂ Gi+1. Now let N be any normal subgroup of G, let k be the largest index such that Nk = N ∩ Gk 6= {e}. Then, {e} = Nk+1 ⊃ [G, Nk] and hence, Nk ⊂ Z(G), where Z(G) is the center of G. Now let N = ker(f). Since the kernel of f restricted to Z(G) is trivial, and ker(f) ∩ Z(G) = N ∩ Z(G) = N is equal to the kernel of f restricted to Z(G), we have that N = {e}, and hence f is injective.