(a) Show that G â H is Abelian if and only if G and H are Abelain ...
MAT301 ASSIGNMENT 6 DUE DATE: AUGUST 6, 2014 AT THE BEGINNING OF YOUR TUTORIAL
Question 1. (a) Show that G ⊕ H is Abelian if and only if G and H are Abelain. Solution. Suppose G and H are Abelian. Given x, y ∈ G ⊕ H, then x = (g, h), y = (a, b) for some g, a ∈ G, h, b ∈ H. Moreover, xy = (g, h)(a, b) = (ga, hb) = (ag, bh) = (a, b)(g, h) = yx Hence, G ⊕ H is Abelian. Conversely, suppose G ⊕ H is Abelian. Let g, a ∈ G, then (g, e), (a, e) ∈ G ⊕ H (where e is the identity element in H), and (ga, e) = (g, e)(a, e) = (a, e)(g, e) = (ag, e) ⇒ ga = ag Hence, G is an Abelian group. A similar proof shows that H is an Abelian group. (b) Prove that G is Abelain if and only if |Inn(G)| = 1. Solution. Suppose G is Abelian. Given φg ∈ Inn(G), we have φg (x) = gxg −1 = xgg −1 = x = φe (x) ⇒ φg = φe (where e is the identity element in G. Hence, Inn(G) = {φe } ⇒ |Inn(G)| = 1. Conversely, suppose |Inn(G)| = 1 ⇒ Inn(G) = {φe }. Given g, h ∈ G, then φg , φh ∈ Inn(G) ⇒ φg = φe = φh ⇒ ghg −1 = φg (h) = φe (h) = h ⇒ gh = hg. Hence, G is Abelian. (c) Suppose G1 , G2 , ..., Gn are groups. Let G=
n M
Gi
i=1
Prove Z(G) =
n M
Z(Gi )
i=1
Solution. Suppose x∈
n M
Z(Gi ) ⇒ x = (g1 , g2 , ..., gn ) for some gi ∈ Z(Gi )
i=1
Given y = (h1 , h2 , ..., hn ) ∈ G, xy = (g1 , g2 , ..., gn )(h1 , h2 , ..., hn ) = (g1 h1 , g2 h2 , ..., gn hn ) = (h1 g1 , h2 g2 , ..., hn gn ) = (h1 , h2 , ..., hn )(g1 , g2 , ..., gn ) = yx That is, x ∈ Z(G). Hence, n M
! ⊆ Z(G)
Z(Gi )
i=1 1
2
DUE DATE: AUGUST 6, 2014 AT THE BEGINNING OF YOUR TUTORIAL
Conversely, suppose x = (g1 , g2 , ..., gn ) ∈ Z(G). For any h1 ∈ G1 , (h1 , e2 , ..., en ) ∈ G (where ei are the identity of Gi ). Hence, (g1 h1 , g2 , ..., gn ) = (g1 , g2 , ..., gn )(h1 , e2 , ..., en ) = (h1 , e2 , ..., en )(g1 , g2 , ..., gn ) = (h1 g1 , g2 , ..., gn ) Therefore, g1 h1 = h1 g1 . Since h1 ∈ G1 was chosen arbitrarily, we have g1 ∈ Z(G1 ). A similar argument shows that gi ∈ Z(Gi ) (replacing h1 by hi and G1 by Gi in the above argument). Then is, x∈
n M
Z(Gi )
i=1
which in turn implies that n M
! Z(Gi )
⊇ Z(G)
i=1
In particular, n M
! Z(Gi )
= Z(G)
i=1
Question 2. Let G be a group and H = {(g, g)| g ∈ G}. Prove H G ⊕ G (H is called the diagonal of G ⊕ G). Solution. Let e be the identity element. Notice that ε = (e, e) is the identity element in G ⊕ G (show this to convince yourselves). In particular, ε ∈ H ⇒ H 6= ∅. Suppose x, y ∈ H ⇒ x = (g, g), y = (h, h) for some g, h ∈ G. Notice that y −1 = (h, h)−1 = (h−1 , h−1 ) (I leave the proof to you). Hence, xy −1 = (g, g)(h−1 , h−1 ) = (gh−1 , gh−1 ) ∈ H Therefore, H G. Question 3. Let 1 a b H = 0 1 0 | a, b ∈ Z3 0 0 1 H is a group of order 9, where the operation is matrix multiplication (with each entry taken mod 3). (a) Show that H is Abelian. Solution. 1 a b 1 c d 1 c+a d+b 1 a+c b+d 1 c d 1 a b 0 1 0 0 1 0 = 0 1 0 = 0 1 0 = 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 (b) Is H isomorphic to Z9 or to Z3 ⊕ Z3 . Solution. 3 1 a b 1 3a 3b 1 0 0 0 1 0 = 0 1 0 = 0 1 0 0 0 1 0 0 1 0 0 1
this shows that every element in H has at most order 3. In particular, H is not cyclic. Hence, H ∼ = Z3 ⊕ Z3 (by theorem in class it must be isomorphic to either Z9 or Z3 ⊕ Z3 ).
MAT301 ASSIGNMENT 6
3
Question 4. Suppose ϕ is an isomorphism from Z3 ⊕ Z5 to Z15 and ϕ(2, 3) = 2. Find the element in (a, b) ∈ Z3 ⊕ Z5 such that ϕ(a, b) = 1 (explain your answer). Note. Since ϕ is an isomorphism, there is only one such element. Solution. 1 ≡ 16 (mod 15) = 8ϕ(2, 3) = ϕ(16 (mod 3), 24 (mod 5)) = ϕ(1, 4). Question 5. Let H and K be subgroups of G. Prove that HK G if and only if HK = KH. Solution. Suppose HK = KH. Notice that e = ee ∈ HK (since H and K are subgroups of G). Given a, b ∈ HK ⇒ a = hk, b = xy for some h, x ∈ H, k, y ∈ K. Let k1 = ky −1 ∈ K. Since KH = HK, there exist c ∈ K, d ∈ H such that hk1 = cd. Let x1 = dx−1 ∈ H. Then, ab−1 = (hk)(xy)−1 = h(ky −1 )x−1 = (hk1 )x−1 = (cd)x−1 = c(dx−1 ) = cx1 ∈ KH = HK Hence, HK G. Conversely, suppose HK G. If kh ∈ KH, then h−1 k −1 ∈ HK ⇒ kh = (h−1 k −1 )−1 ∈ HK (since HK G, in particular, closed under inverses). Hence, KH ⊆ HK. If hk ∈ HK ⇒ (hk)−1 ∈ HK (since HK G) ⇒ k −1 h−1 ∈ HK. In particular, k −1 h−1 = ab for some a ∈ H, b ∈ K ⇒ hk = (k −1 h−1 )−1 = (ab)−1 = b−1 a−1 ∈ KH. Therefore, HK ⊆ KH. Hence, HK = KH. Question 6. (a) Suppose |G| = n < ∞, where n is an even number. Prove that any subgroup H of G with [G : H] = 2 is normal. Solution. Suppose [G : H] = 2. Then this tells us that the set of left cosets of H in G and the set of right cosest of H in G has two elements. Given a ∈ G, we consider two cases; Case 1. a ∈ H ⇒ aH = H = Ha. Case 2. a ∈ / H ⇒ aH 6= H and Ha 6= H. In particular, {H, aH} is the set of left cosets of H in G, and {H, Ha} is the set of right cosets of H in G. Since the cosets of H (right or left) partition G, aH 6= H ⇒ aH ∩ H = ∅ ⇒ aH ⊆ Ha, and Ha 6= H ⇒ Ha ∩ H = ∅ ⇒ Ha ⊆ aH. Hence, aH = Ha, as desired. Therefore, H E G. (b) Conclude that An E Sn (for n ≥ 2). Solution. Since [Sn : An ] = 2, by part (a) An E Sn . (c) Show that the converse of part (a) is false. That is, give an example of a normal subgroup H of a finite group G such that [G : H] 6= 2. Solution. Let G = Z10 and H = e (the trivial subgroup). Then H E G, however, [G : H] = 10. There are many other examples, in fact, you can take G to be any finite group with more than 2 elements and take H to be the trivial subgroup of G. Also there are examples of infinite groups with subgroups that are normal but do not have index 2 (you should try to come up with such an example). Question 7. (a) Suppose G is an Abelian group. Prove that G/H is an Abelian group for all H E G (You do not have to show that G/H is a group, just show it is Abelian). Solution. Suppose G is an Abelian group. Given x, y ∈ G/H, then x = gH, y = bH for some g, b ∈ G. Then xy = (gH)(bH) = (gb)H = (bg)H = (bH)(gH) = yx Hence, G/H is an Abelian group. (b) Suppose G is cyclic. Prove that G/H is cyclic for all H E G (You do not have to show G/H is a group, just show it is cyclic).
4
DUE DATE: AUGUST 6, 2014 AT THE BEGINNING OF YOUR TUTORIAL
Solution. Suppose G is a cyclic group, say G = hgi. Notice that hgHi ⊆ G/H (since G/H is a group and gH ∈ G/H). Conversely, given x ∈ G/H, then x = aH for some a ∈ G. Moreover, a = g n for some n ∈ Z. In particular, x = aH = (g n )H = (gH)n ∈ hgHi. Hence, G/H ⊆ hgHi. Therefore, G/H = hgHi (that is, G/H is cyclic). Question 8. Suppose G is a group and [G : Z(G)] = 4. Prove that G/Z(G) ∼ = Z2 ⊕ Z2 . Solution. By theorem in class we have G/Z(G) ∼ = Z4 or Z2 ⊕ Z2 . Assume G/Z(G) ∼ = Z4 ⇒ G/Z(G) is cyclic ⇒ G is Abelian (by another theorem we have done in class) ⇒ G = Z(G) ⇒ [G : Z(G)] = 1, a contradiction. Hence, G/Z(G) ∼ = Z2 ⊕ Z2 . Question 9. Let G be a group and let G0 be the subgroup of G generated by the set S = {x−1 y −1 xy| x, y ∈ G} (note that S = {e} if and only if G is Abelian). Definition. Let [x, y] = x−1 y −1 xy (this is called the commutator of x and y). (a) Prove that G0 E G. Solution. It suffices to show g −1 G0 g ⊆ G0 for all g ∈ G. Pick any g ∈ G. Given x ∈ G0 , x = [x1 , y1 ]...[xn , yn ] for some xi , yi ∈ G. Moreover, −1 −1 −1 −1 −1 −1 −1 −1 g −1 [xi , yi ]g = g −1 x−1 i yi xi yi g = (xi yi xi yi )(yi xi yi xi )g xi yi xi yi g −1 −1 −1 −1 −1 −1 −1 = (x−1 i yi xi yi )((xi yi xi yi ) g xi yi xi yi g) −1 0 = [xi , yi ][x−1 i yi xi yi , g] ∈ G
In particular, x ∈ G0 (since G0 is a group). Since g ∈ G and x ∈ G0 were chosen arbitrarily, G0 E G. (b) Prove that G/G0 is Abelian. Solution. Define π : G → G/G0 via π(g) = gG0 . Then π is a surjective homomorphism (I leave the proof of this fact to you). Notice that for any x, y ∈ G, π([x, y]) = [x, y]G0 = G0 (the identity element in G/G0 ) ⇒ π(x)π(y) = π(y)π(x). Given g, h ∈ G/G0 , then g = xG0 , h = yG0 for some x, y ∈ G. Moreover, gh = (xG0 )(yG0 ) = π(x)π(y) = π(y)π(x) = (yG0 )(xG0 ) = hg Hence G/G0 is an Abelian group. (c) Suppose N E G such that G/N is Abelain. Prove that G0 N . Solution. Define π : G → G/N via π(g) = gN (similar map to part (b)). Then, π is a surjective homomorphism. Given g, h ∈ G, π(g), π(h) ∈ G/N . Since G/N is Abelian, π(g)π(h) = π(h)π(g) ⇒ π(gh) = π(hg) ⇒ (gh)N = (hg)N ⇒ (hg)−1 gh ∈ N ⇒ g −1 h−1 gh ∈ N ⇒ [g, h] ∈ N . This shows that for any g, h ∈ G, [g, h] ∈ N . In particular, S ⊆ N . Since N G (in particular, it is closed under multiplication and inverses), we get G0 ⊆ N . Moreover, since G0 is a group (after all G0 being a subgroup means it is a group), we have G0 N . (d) Prove that if G0 H G, then H E G. Solution. Given g ∈ G, and h ∈ H, we have [g −1 , h−1 ] ∈ G0 H ⇒ [g −1 , h−1 ] ∈ H. This implies that ghg −1 = ghg −1 h−1 h = [g −1 , h−1 ]h ∈ H Since g ∈ G, and h ∈ H were chosen arbitrarily, we have gHg −1 ⊆ H for all g ∈ G. Hence, H E G. Question 10. Give an example of a group G with non-isomorphic normal subgroups H and K such that G/H and G/K are isomorphic. Solution. Let G = Z4 ⊕ Z2 . Notice that G is an Abelian group, hence, every subgroup of it is a normal subgroup. Let H = Z2 ⊕ Z2 , K = Z4 ⊕ {e} (where e is the identity element in G). Notice that K ∼ = Z4 H. However, G/H = (Z4 ⊕ Z2 )/Z2 ⊕ Z2 ) ∼ = Z2 ∼ = (Z4 ⊕ Z2 )/Z4 ⊕ {e} = G/K
Question 1. (a) Show that G ⊕ H is Abelian if and only if G and H are Abelain. Solution. Suppose G and H are Abelian. Given x, y ∈ G ⊕ H, then x = (g, h), y = (a, b) for some g, a ∈ G, h, b ∈ H. Moreover, xy = (g, h)(a, b) = (ga, hb) = (ag, bh) = (a, b)(g, h) = yx Hence, G ⊕ H is Abelian. Conversely, suppose G ⊕ H is Abelian. Let g, a ∈ G, then (g, e), (a, e) ∈ G ⊕ H (where e is the identity element in H), and (ga, e) = (g, e)(a, e) = (a, e)(g, e) = (ag, e) ⇒ ga = ag Hence, G is an Abelian group. A similar proof shows that H is an Abelian group. (b) Prove that G is Abelain if and only if |Inn(G)| = 1. Solution. Suppose G is Abelian. Given φg ∈ Inn(G), we have φg (x) = gxg −1 = xgg −1 = x = φe (x) ⇒ φg = φe (where e is the identity element in G. Hence, Inn(G) = {φe } ⇒ |Inn(G)| = 1. Conversely, suppose |Inn(G)| = 1 ⇒ Inn(G) = {φe }. Given g, h ∈ G, then φg , φh ∈ Inn(G) ⇒ φg = φe = φh ⇒ ghg −1 = φg (h) = φe (h) = h ⇒ gh = hg. Hence, G is Abelian. (c) Suppose G1 , G2 , ..., Gn are groups. Let G=
n M
Gi
i=1
Prove Z(G) =
n M
Z(Gi )
i=1
Solution. Suppose x∈
n M
Z(Gi ) ⇒ x = (g1 , g2 , ..., gn ) for some gi ∈ Z(Gi )
i=1
Given y = (h1 , h2 , ..., hn ) ∈ G, xy = (g1 , g2 , ..., gn )(h1 , h2 , ..., hn ) = (g1 h1 , g2 h2 , ..., gn hn ) = (h1 g1 , h2 g2 , ..., hn gn ) = (h1 , h2 , ..., hn )(g1 , g2 , ..., gn ) = yx That is, x ∈ Z(G). Hence, n M
! ⊆ Z(G)
Z(Gi )
i=1 1
2
DUE DATE: AUGUST 6, 2014 AT THE BEGINNING OF YOUR TUTORIAL
Conversely, suppose x = (g1 , g2 , ..., gn ) ∈ Z(G). For any h1 ∈ G1 , (h1 , e2 , ..., en ) ∈ G (where ei are the identity of Gi ). Hence, (g1 h1 , g2 , ..., gn ) = (g1 , g2 , ..., gn )(h1 , e2 , ..., en ) = (h1 , e2 , ..., en )(g1 , g2 , ..., gn ) = (h1 g1 , g2 , ..., gn ) Therefore, g1 h1 = h1 g1 . Since h1 ∈ G1 was chosen arbitrarily, we have g1 ∈ Z(G1 ). A similar argument shows that gi ∈ Z(Gi ) (replacing h1 by hi and G1 by Gi in the above argument). Then is, x∈
n M
Z(Gi )
i=1
which in turn implies that n M
! Z(Gi )
⊇ Z(G)
i=1
In particular, n M
! Z(Gi )
= Z(G)
i=1
Question 2. Let G be a group and H = {(g, g)| g ∈ G}. Prove H G ⊕ G (H is called the diagonal of G ⊕ G). Solution. Let e be the identity element. Notice that ε = (e, e) is the identity element in G ⊕ G (show this to convince yourselves). In particular, ε ∈ H ⇒ H 6= ∅. Suppose x, y ∈ H ⇒ x = (g, g), y = (h, h) for some g, h ∈ G. Notice that y −1 = (h, h)−1 = (h−1 , h−1 ) (I leave the proof to you). Hence, xy −1 = (g, g)(h−1 , h−1 ) = (gh−1 , gh−1 ) ∈ H Therefore, H G. Question 3. Let 1 a b H = 0 1 0 | a, b ∈ Z3 0 0 1 H is a group of order 9, where the operation is matrix multiplication (with each entry taken mod 3). (a) Show that H is Abelian. Solution. 1 a b 1 c d 1 c+a d+b 1 a+c b+d 1 c d 1 a b 0 1 0 0 1 0 = 0 1 0 = 0 1 0 = 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 (b) Is H isomorphic to Z9 or to Z3 ⊕ Z3 . Solution. 3 1 a b 1 3a 3b 1 0 0 0 1 0 = 0 1 0 = 0 1 0 0 0 1 0 0 1 0 0 1
this shows that every element in H has at most order 3. In particular, H is not cyclic. Hence, H ∼ = Z3 ⊕ Z3 (by theorem in class it must be isomorphic to either Z9 or Z3 ⊕ Z3 ).
MAT301 ASSIGNMENT 6
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Question 4. Suppose ϕ is an isomorphism from Z3 ⊕ Z5 to Z15 and ϕ(2, 3) = 2. Find the element in (a, b) ∈ Z3 ⊕ Z5 such that ϕ(a, b) = 1 (explain your answer). Note. Since ϕ is an isomorphism, there is only one such element. Solution. 1 ≡ 16 (mod 15) = 8ϕ(2, 3) = ϕ(16 (mod 3), 24 (mod 5)) = ϕ(1, 4). Question 5. Let H and K be subgroups of G. Prove that HK G if and only if HK = KH. Solution. Suppose HK = KH. Notice that e = ee ∈ HK (since H and K are subgroups of G). Given a, b ∈ HK ⇒ a = hk, b = xy for some h, x ∈ H, k, y ∈ K. Let k1 = ky −1 ∈ K. Since KH = HK, there exist c ∈ K, d ∈ H such that hk1 = cd. Let x1 = dx−1 ∈ H. Then, ab−1 = (hk)(xy)−1 = h(ky −1 )x−1 = (hk1 )x−1 = (cd)x−1 = c(dx−1 ) = cx1 ∈ KH = HK Hence, HK G. Conversely, suppose HK G. If kh ∈ KH, then h−1 k −1 ∈ HK ⇒ kh = (h−1 k −1 )−1 ∈ HK (since HK G, in particular, closed under inverses). Hence, KH ⊆ HK. If hk ∈ HK ⇒ (hk)−1 ∈ HK (since HK G) ⇒ k −1 h−1 ∈ HK. In particular, k −1 h−1 = ab for some a ∈ H, b ∈ K ⇒ hk = (k −1 h−1 )−1 = (ab)−1 = b−1 a−1 ∈ KH. Therefore, HK ⊆ KH. Hence, HK = KH. Question 6. (a) Suppose |G| = n < ∞, where n is an even number. Prove that any subgroup H of G with [G : H] = 2 is normal. Solution. Suppose [G : H] = 2. Then this tells us that the set of left cosets of H in G and the set of right cosest of H in G has two elements. Given a ∈ G, we consider two cases; Case 1. a ∈ H ⇒ aH = H = Ha. Case 2. a ∈ / H ⇒ aH 6= H and Ha 6= H. In particular, {H, aH} is the set of left cosets of H in G, and {H, Ha} is the set of right cosets of H in G. Since the cosets of H (right or left) partition G, aH 6= H ⇒ aH ∩ H = ∅ ⇒ aH ⊆ Ha, and Ha 6= H ⇒ Ha ∩ H = ∅ ⇒ Ha ⊆ aH. Hence, aH = Ha, as desired. Therefore, H E G. (b) Conclude that An E Sn (for n ≥ 2). Solution. Since [Sn : An ] = 2, by part (a) An E Sn . (c) Show that the converse of part (a) is false. That is, give an example of a normal subgroup H of a finite group G such that [G : H] 6= 2. Solution. Let G = Z10 and H = e (the trivial subgroup). Then H E G, however, [G : H] = 10. There are many other examples, in fact, you can take G to be any finite group with more than 2 elements and take H to be the trivial subgroup of G. Also there are examples of infinite groups with subgroups that are normal but do not have index 2 (you should try to come up with such an example). Question 7. (a) Suppose G is an Abelian group. Prove that G/H is an Abelian group for all H E G (You do not have to show that G/H is a group, just show it is Abelian). Solution. Suppose G is an Abelian group. Given x, y ∈ G/H, then x = gH, y = bH for some g, b ∈ G. Then xy = (gH)(bH) = (gb)H = (bg)H = (bH)(gH) = yx Hence, G/H is an Abelian group. (b) Suppose G is cyclic. Prove that G/H is cyclic for all H E G (You do not have to show G/H is a group, just show it is cyclic).
4
DUE DATE: AUGUST 6, 2014 AT THE BEGINNING OF YOUR TUTORIAL
Solution. Suppose G is a cyclic group, say G = hgi. Notice that hgHi ⊆ G/H (since G/H is a group and gH ∈ G/H). Conversely, given x ∈ G/H, then x = aH for some a ∈ G. Moreover, a = g n for some n ∈ Z. In particular, x = aH = (g n )H = (gH)n ∈ hgHi. Hence, G/H ⊆ hgHi. Therefore, G/H = hgHi (that is, G/H is cyclic). Question 8. Suppose G is a group and [G : Z(G)] = 4. Prove that G/Z(G) ∼ = Z2 ⊕ Z2 . Solution. By theorem in class we have G/Z(G) ∼ = Z4 or Z2 ⊕ Z2 . Assume G/Z(G) ∼ = Z4 ⇒ G/Z(G) is cyclic ⇒ G is Abelian (by another theorem we have done in class) ⇒ G = Z(G) ⇒ [G : Z(G)] = 1, a contradiction. Hence, G/Z(G) ∼ = Z2 ⊕ Z2 . Question 9. Let G be a group and let G0 be the subgroup of G generated by the set S = {x−1 y −1 xy| x, y ∈ G} (note that S = {e} if and only if G is Abelian). Definition. Let [x, y] = x−1 y −1 xy (this is called the commutator of x and y). (a) Prove that G0 E G. Solution. It suffices to show g −1 G0 g ⊆ G0 for all g ∈ G. Pick any g ∈ G. Given x ∈ G0 , x = [x1 , y1 ]...[xn , yn ] for some xi , yi ∈ G. Moreover, −1 −1 −1 −1 −1 −1 −1 −1 g −1 [xi , yi ]g = g −1 x−1 i yi xi yi g = (xi yi xi yi )(yi xi yi xi )g xi yi xi yi g −1 −1 −1 −1 −1 −1 −1 = (x−1 i yi xi yi )((xi yi xi yi ) g xi yi xi yi g) −1 0 = [xi , yi ][x−1 i yi xi yi , g] ∈ G
In particular, x ∈ G0 (since G0 is a group). Since g ∈ G and x ∈ G0 were chosen arbitrarily, G0 E G. (b) Prove that G/G0 is Abelian. Solution. Define π : G → G/G0 via π(g) = gG0 . Then π is a surjective homomorphism (I leave the proof of this fact to you). Notice that for any x, y ∈ G, π([x, y]) = [x, y]G0 = G0 (the identity element in G/G0 ) ⇒ π(x)π(y) = π(y)π(x). Given g, h ∈ G/G0 , then g = xG0 , h = yG0 for some x, y ∈ G. Moreover, gh = (xG0 )(yG0 ) = π(x)π(y) = π(y)π(x) = (yG0 )(xG0 ) = hg Hence G/G0 is an Abelian group. (c) Suppose N E G such that G/N is Abelain. Prove that G0 N . Solution. Define π : G → G/N via π(g) = gN (similar map to part (b)). Then, π is a surjective homomorphism. Given g, h ∈ G, π(g), π(h) ∈ G/N . Since G/N is Abelian, π(g)π(h) = π(h)π(g) ⇒ π(gh) = π(hg) ⇒ (gh)N = (hg)N ⇒ (hg)−1 gh ∈ N ⇒ g −1 h−1 gh ∈ N ⇒ [g, h] ∈ N . This shows that for any g, h ∈ G, [g, h] ∈ N . In particular, S ⊆ N . Since N G (in particular, it is closed under multiplication and inverses), we get G0 ⊆ N . Moreover, since G0 is a group (after all G0 being a subgroup means it is a group), we have G0 N . (d) Prove that if G0 H G, then H E G. Solution. Given g ∈ G, and h ∈ H, we have [g −1 , h−1 ] ∈ G0 H ⇒ [g −1 , h−1 ] ∈ H. This implies that ghg −1 = ghg −1 h−1 h = [g −1 , h−1 ]h ∈ H Since g ∈ G, and h ∈ H were chosen arbitrarily, we have gHg −1 ⊆ H for all g ∈ G. Hence, H E G. Question 10. Give an example of a group G with non-isomorphic normal subgroups H and K such that G/H and G/K are isomorphic. Solution. Let G = Z4 ⊕ Z2 . Notice that G is an Abelian group, hence, every subgroup of it is a normal subgroup. Let H = Z2 ⊕ Z2 , K = Z4 ⊕ {e} (where e is the identity element in G). Notice that K ∼ = Z4 H. However, G/H = (Z4 ⊕ Z2 )/Z2 ⊕ Z2 ) ∼ = Z2 ∼ = (Z4 ⊕ Z2 )/Z4 ⊕ {e} = G/K
