Math 110A HW §3.5 – Solutions
7. Define φ : Z → H by φ(n) =
1 n
.
0 1
We first prove one to one. Let n, m ∈ Z such that φ(n) = φ(m). So
1 n
0 1
=
1 m
0 1
which implies n = m. Thus φ is one to one.
1 n
1 n
. Thus φ is onto.
∈ H, then n ∈ Z and φ(n) =
We now prove onto. Let
0 1
0 1
Finally let’s check that φ preserves the group structure.
1 n+m
φ(n + m) =
0
1
1 n
1 m
=
0 1
0 1
= φ(n)φ(m)
Thus φ is an isomorphism.
12. Let φ : R → R+ given by φ(x) = 10x. Prove φ is an isomorphism.
Proof. Let a, b ∈ R such that φ(a) = φ(b). Thus 10a = 10b. Hence a = b.
Let y ∈ R+ . Then log y ∈ R. Moreover φ(log y) = 10log y = y. Thus φ is onto.
Let a, b ∈ R. Then φ(a + b) = 10a+b = 10a · 10b = φ(a)φ(b).
Thus φ is an isomorphism.
14. Let φ : C∗ → C∗ defined by φ(a + bi) = a − bi. Prove that φ is an automorphism.
Proof. Let’s first prove that φ preserves the group structure. Let a + bi, c + di ∈ C. Then we
have the following.
φ(a + bi)φ(c + di) = (a − bi)(c − di)
= (ac − bd) − (ad + bc)i
= φ (ac − bd) + (ad + bc)i
= φ (a + bi)(c + di)
Now let’s prove one-to-one. Suppose φ(a + bi) = φ(c + di). Thus a − bi = c − di and hence
a = c and b = d. Therefore a + bi = c + di. Thus φ is one-to-one.
Now we will prove onto. Let a + bi ∈ C. Then a − bi ∈ C and φ(a − bi) = a + bi. Therefore
φ is onto.
Hence φ is an automorphism on C.
16. Suppose (m, n) = 1 and let φ : Zn → Zn be defined by φ([a]) = m[a]. Prove or disprove that
φ is an automorphism.
Proof. We first need to check well-defined. Suppose [a] = [b]. Then a ≡ b(mod n) and hence
ma ≡ mb(mod n). Thus m[a] = m[b]. Therefore φ is well-defined.
Let’s check that φ preserves group structure.
φ([a] + [b]) = φ([a + b]) = m[a + b] = m([a] + [b]) = m[a] + m[b] = φ([a]) + φ([b])
Now let’s prove one-to-one. Suppose φ([a]) = φ([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 φ is one-to-one.
Now let’s prove onto. Let [a] ∈ Zn . Since (m, n) = 1 there are integers c and d such that
1 = mc + nd. Multiplying this by a gives a = mac + nad and hence a ≡ mac(mod n). Thus
[a] = [mac]. Therefore we have the following: φ([ac]) = m[ac] = [mac] = [a]. Thus φ is onto.
Therefore φ is an automorphism.
18. For each a in the group G, define a mapping ta : G → G by ta (x) = axa−1 . Prove that ta is
an automorphism.
Proof. We first prove one-to-one. Suppose x, y ∈ G such that ta (x) = ta (y). Thus we have
the following.
ta (x) = ta (y)
axa−1 = aya−1
xa−1 = ya−1
x=y
Thus ta is one-to-one.
We now prove onto. Let x ∈ G. Then a−1 xa ∈ G and ta (a−1 xa) = a(a−1xa)a−1 = x. Thus
ta is onto.
Finally we prove ta preserves the group structure.
ta (xy) = axya−1 = axa−1 aya−1 = 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 is
abelian.
Proof. Assume G and H are isomorphic, then there exists an isomoprhism φ : G → H.
Assume G is abelian. Let x, y ∈ H. Since φ is onto there exists a, b ∈ G such that φ(a) = x
and φ(b) = y. Then we have xy = φ(a)φ(b) = φ(ab) since φ preserves group structure. Since
G is abelian, φ(ab) = φ(ba). Finally, since φ preserves group structure φ(ba) = φ(b)φ(a) = yx.
Thus we have shown that xy = yx. Hence H is abelian.
Assume H is abelian. Let a, b ∈ G. Then φ(a), φ(b) ∈ H and since H is abelian φ(a)φ(b) =
φ(b)φ(a). Thus we have φ(ab) = φ(a)φ(b) = φ(b)φ(a) = φ(ba). Now, since φ is one-to-one
and φ(ab) = φ(ba), then we have that ab = ba. Thus G is abelian.
33. If G and G0 are groups and φ : G → G0 is an isomorphism, prove that a and φ(a) have the
same order for any a ∈ G.
Proof. Let |a| = m, so am = e. Then φ(a)m = φ(am ) = φ(e) = e0 . Where e0 is the identity
element of G0 .
Suppose there is some positive integer j such that φ(a)j = e0 . Then e0 = φ(a)j = φ(aj ), but
φ(e) = e0 . Since φ is one-to-one, aj = e. However, |a| = m, hence m ≤ j.
Therefore we have shown that φ(a)m = e0 and m is the smallest positive such power. Hence
|φ(a)| = m.
34. Suppose that φ is an isomorphism from the group G to the group G0.
(a) Prove that if H is a subgroup of G, then φ(H) is a subgroup of G0 .
Proof. Since H is a subgroup of G, e ∈ H. Moreover φ(e) = e0. Thus e0 ∈ φ(H).
Let x, y ∈ φ(H). So there exists a, b ∈ H such that φ(a) = x and φ(b) = y. Thus
xy = φ(a)φ(b) = φ(ab). Since H is a group ab ∈ H. Thus xy = φ(ab) ∈ φ(H).
Let z ∈ φ(H). So there exists c ∈ H such that φ(c) = z. Thus z −1 = φ(c)−1 = φ(c−1 ).
Since H is a group a−1 ∈ H. Thus z −1 = φ(c−1) ∈ φ(H).
Therefore by the subgroup test, φ(H) is a subgroup of G0 .
(b) Suppose that φ is an isomorphism from G to G0. Prove that if K is a subgroup of G0,
then φ−1 (K) is a subgroup of G.
Proof. Since φ(e) = e0 ∈ K, e ∈ φ−1 (K). Thus φ−1 (K) 6= ∅.
Let a, b ∈ φ−1 (K). So φ(a), φ(b) ∈ K. Since K is a subgroup of G0, φ(a)φ(b) ∈ K.
Therefore we have the following.
φ(ab) = φ(a)φ(b) ∈ K
Thus ab ∈ φ−1 (K). Hence φ−1 (K) is closed.
Let a ∈ φ−1 (K). So φ(a) ∈ K. Since K is a subgroup of G0 , φ(a)−1 ∈ K. Therefore
φ(a−1 ) = φ(a)−1 ∈ K. Thus a−1 ∈ φ−1 (K).
Therefore φ−1 (K) is a subgroup of G0.