Math 110A HW 3.5 Solutions

7. Define : Z H by (n) =[

1 n0 1

].

We first prove one to one. Let n, m Z such that (n) = (m). So[

1 n0 1

]=

[1 m0 1

]

which implies n = m. Thus is one to one.

We now prove onto. Let[

1 n0 1

] H , then n Z and (n) =

[1 n0 1

]. Thus is onto.

Finally lets check that preserves the group structure.

(n + m) =[

1 n + m0 1

]

=[

1 n0 1

] [1 m0 1

]

= (n)(m)

Thus is an isomorphism.

12. Let : R R+ given by (x) = 10x. Prove is an isomorphism.

Proof. Let a, b R such that (a) = (b). Thus 10a = 10b. Hence a = b.Let y R+. Then log y R. Moreover (log y) = 10logy = y. Thus is onto.Let a, b R. Then (a + b) = 10a+b = 10a 10b = (a)(b).Thus is an isomorphism.

14. Let : C C defined by (a + bi) = a bi. Prove that is an automorphism.

Proof. Lets first prove that preserves the group structure. Let a + bi, c+ di C. Then wehave the following.

(a + bi)(c + di) = (a bi)(c di)= (ac bd) (ad + bc)i

= ((ac bd) + (ad + bc)i

)

= ((a + bi)(c + di)

)

Now lets prove one-to-one. Suppose (a + bi) = (c + di). Thus a bi = c di and hencea = c and b = d. Therefore a + bi = c + di. Thus is one-to-one.

Now we will prove onto. Let a + bi C. Then a bi C and (a bi) = a + bi. Therefore is onto.

Hence is an automorphism on C.

16. Suppose (m, n) = 1 and let : Zn Zn be defined by ([a]) = m[a]. Prove or disprove that is an automorphism.

Proof. We first need to check well-defined. Suppose [a] = [b]. Then a b(mod n) and hencema mb(mod n). Thus m[a] = m[b]. Therefore is well-defined.Lets check that preserves group structure.

([a] + [b]) = ([a + b]) = m[a + b] = m([a] + [b]) = m[a] + m[b] = ([a]) + ([b])

Now lets prove one-to-one. Suppose ([a]) = ([b]). Then m[a] = m[b]. So ma mb(mod n).Since (m, n) = 1 this implies that a b(mod n). Thus [a] = [b] and hence is one-to-one.Now lets prove onto. Let [a] Zn. Since (m, n) = 1 there are integers c and d such that1 = mc + nd. Multiplying this by a gives a = mac + nad and hence a mac(mod n). Thus[a] = [mac]. Therefore we have the following: ([ac]) = m[ac] = [mac] = [a]. Thus is onto.

Therefore is an automorphism.

18. For each a in the group G, define a mapping ta : G G by ta(x) = axa1. Prove that ta isan automorphism.

Proof. We first prove one-to-one. Suppose x, y G such that ta(x) = ta(y). Thus we havethe following.

ta(x) = ta(y)

axa1 = aya1

xa1 = ya1

x = y

Thus ta is one-to-one.

We now prove onto. Let x G. Then a1xa G and ta(a1xa) = a(a1xa)a1 = x. Thusta is onto.

Finally we prove ta preserves the group structure.

ta(xy) = axya1 = axa1aya1 = ta(x)ta(y)

Thus ta is an isomorphism.

28. Suppose that G and H are isomorphic groups. Prove that G is abelian if and only if H isabelian.

Proof. Assume G and H are isomorphic, then there exists an isomoprhism : G H .Assume G is abelian. Let x, y H . Since is onto there exists a, b G such that (a) = xand (b) = y. Then we have xy = (a)(b) = (ab) since preserves group structure. SinceG is abelian, (ab) = (ba). Finally, since preserves group structure (ba) = (b)(a) = yx.Thus we have shown that xy = yx. Hence H is abelian.

Assume H is abelian. Let a, b G. Then (a), (b) H and since H is abelian (a)(b) =(b)(a). Thus we have (ab) = (a)(b) = (b)(a) = (ba). Now, since is one-to-oneand (ab) = (ba), then we have that ab = ba. Thus G is abelian.

33. If G and G are groups and : G G is an isomorphism, prove that a and (a) have thesame order for any a G.

Proof. Let |a| = m, so am = e. Then (a)m = (am) = (e) = e. Where e is the identityelement of G.

Suppose there is some positive integer j such that (a)j = e. Then e = (a)j = (aj), but(e) = e. Since is one-to-one, aj = e. However, |a| = m, hence m j.Therefore we have shown that (a)m = e and m is the smallest positive such power. Hence|(a)| = m.

34. Suppose that is an isomorphism from the group G to the group G.

(a) Prove that if H is a subgroup of G, then (H) is a subgroup of G.

Proof. Since H is a subgroup of G, e H . Moreover (e) = e. Thus e (H).Let x, y (H). So there exists a, b H such that (a) = x and (b) = y. Thusxy = (a)(b) = (ab). Since H is a group ab H . Thus xy = (ab) (H).Let z (H). So there exists c H such that (c) = z. Thus z1 = (c)1 = (c1).Since H is a group a1 H . Thus z1 = (c1) (H).Therefore by the subgroup test, (H) is a subgroup of G.

(b) Suppose that is an isomorphism from G to G. Prove that if K is a subgroup of G,then 1(K) is a subgroup of G.

Proof. Since (e) = e K, e 1(K). Thus 1(K) 6= .Let a, b 1(K). So (a), (b) K. Since K is a subgroup of G, (a)(b) K.Therefore we have the following.

(ab) = (a)(b) K

Thus ab 1(K). Hence 1(K) is closed.Let a 1(K). So (a) K. Since K is a subgroup of G, (a)1 K. Therefore(a1) = (a)1 K. Thus a1 1(K).Therefore 1(K) is a subgroup of G.