MATH 3005 Homework Solution
Han-Bom Moon
Homework 6 Solution
Chapter 6.
1. Find an isomorphism from the group of integers under addition to the group of
even integers under addition.
Let 2Z be the set of all even integers. Define a map φ : Z → 2Z as φ(n) = 2n.
We claim that φ is an isomorphism. φ(n) = φ(m) ⇒ 2n = 2m ⇒ n = m so it is
one-to-one. For any even integer 2k, φ(k) = 2k thus it is onto. Also
φ(n + m) = 2(n + m) = 2n + 2m = φ(n) + φ(m),
so it has the operation preserving property.
2. Find Aut(Z).
Note that Z = 1 , a cyclic group generated by 1. There are two generators, 1
and −1. Because an automorphism φ of a cyclic group sends a generator to a
generator, φ(1) = 1 or φ(1) = −1. Because φ(m · 1) = mφ(1), for the former case
we have the identity map, and for the latter case, we have φ(x) = −x. Therefore
Aut(Z) = {id, φ} where φ(x) = −x.
4. Show that U (8) is not isomorphic to U (10).
U (10) = {1, 3, 7, 9} is a cyclic group generated by 3. So 3 is an element of order
4. But all non-identity elements of U (8) = {1, 3, 5, 7} have order 2, so there is no
element of order 4. Therefore they are not isomorphic to each other.
5. Show that U (8) is isomorphic to U (12).
U (8) = {1, 3, 5, 7} and U (12) = {1, 5, 7, 11}. Take a bijective map φ : U (8) →
U (12) defined by φ(1) = 1, φ(3) = 11, φ(5) = 5, and φ(7) = 7. We claim that it has
the operation preserving property. Because U (8) is an Abelian group, it suffices
to check followings:
φ(32 ) = φ(1) = 1 = 112 = φ(3)2 ,
φ(52 ) = φ(1) = 1 = 52 = φ(5)2 ,
φ(72 ) = φ(1) = 1 = 72 = φ(7)2 ,
φ(3 · 5) = φ(7) = 7 = 11 · 5 = φ(3) · φ(5),
φ(3 · 7) = φ(5) = 5 = 11 · 7 = φ(3) · φ(7),
φ(5 · 7) = φ(3) = 11 = 5 · 7 = φ(5) · φ(7).
In general, you may show that any bijective map ψ : U (8) → U (12) with ψ(1) = 1
is an isomorphism.
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MATH 3005 Homework Solution
Han-Bom Moon
10. Let G be a group. Prove that the mapping α(g) = g −1 for all g in G is an automorphism if and only if G is Abelian.
If α is an isomorphism, for any two elements x, y ∈ G,
y −1 x−1 = (xy)−1 = α(xy) = α(x)α(y) = x−1 y −1
So xy = (x−1 )−1 (y −1 )−1 = (y −1 x−1 )−1 = (x−1 y −1 )−1 = (y −1 )−1 (x−1 )−1 = yx,
and G is Abelian.
Suppose that G is Abelian. α : G → G is a bijective function, because α itself is
the inverse function of α. Moreover, because
α(xy) = α(yx) = (yx)−1 = x−1 y −1 = α(x)α(y),
it has the operation preserving property. So α is an isomorphism.
14. Find Aut(Z6 ).
Z6 is a cyclic group generated by 1. There are two generators, 1, 5. So for an
isomorphism φ : Z6 → Z6 , φ(1) = 1 or φ(1) = 5 = −1. We have two such
isomorphisms: the identity map and φ(x) = −x. Therefore Aut(Z6 ) = {id, φ}
where φ(x) = −x.
15. If G is a group, prove that Aut(G) and Inn(G) are groups.
Both sets are subsets of SG , the permutation group of the set G. Because id ∈
Aut(G) and id = φe ∈ Inn(G), we may apply a subgroup test.
Step 1. Aut(G). Suppose that α, β ∈ Aut(G). Then αβ : G → G and α−1 are
elements of SG , so they are bijective. Therefore it is sufficient to show the operation preserving property. For any x, y ∈ G, αβ(xy) = α(β(xy)) = α(β(x)β(y)) =
α(β(x))α(β(y)) = αβ(x)αβ(y). Therefore αβ ∈ Aut(G).
Suppose that α(a) = x, α(b) = y. Then α(ab) = α(a)α(b) = xy. Thus α−1 (xy) =
ab = α−1 (x)α−1 (y) and α−1 has the operation preserving property as well. Therefore α−1 ∈ Aut(G). By the subgroup test 1, Aut(G) ≤ SG .
Step 2. Inn(G). Let φa , φb ∈ Inn(G). Then φa φb (x) = φa (φb (x)) = φa (bxb−1 ) =
abxb−1 a−1 = (ab)x(ab)−1 = φab (x). So φa φb = φab ∈ Inn(G).
Note that φa−1 φa (x) = φa−1 (φa (x)) = φa−1 (axa−1 ) = a−1 axa−1 (a−1 )−1 = a−1 axa−1 a =
x and φa φa−1 x = φa (φa−1 (x)) = φa (a−1 xa) = aa−1 xaa−1 = x. So φ−1
a = φa−1 ∈
Inn(G). By the subgroup test 1, Inn(G) ≤ SG .
24. Suppose that φ : Z20 → Z20 is an automorphism and φ(5) = 5. What are the
possibilities for φ(x)?
Because an automorphism φ maps a generator to a generator, φ(1) is one of
1, 3, 7, 9, 11, 13, 17, 19. Because φ(5) = φ(5·1) = 5φ(1) = 5 in Z20 , the only possible
φ(1) are 1, 9, 13, 17. Therefore φ(x) = x, 9x, 13x, or 17x.
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MATH 3005 Homework Solution
1 a
0 1
is replaced by R?
28. The group
Let G =
1 a
0 1
a∈Z
1 a
0 1
1 b
0 1
is isomorphic to what familiar group? What if Z
a ∈ Z . We claim that G is isomorphic to an additive group
Z. Define φ : G → Z as φ(
φ(
Han-Bom Moon
1 a
0 1
) = φ(
) = a. Obviously it is a bijection map.
1 a+b
0
1
) = a + b = φ(
1 a
0 1
) + φ(
1 a
0 1
).
Therefore φ has the operation preserving property, so it is an isomorphism.
If we replace Z by R, then the group is isomorphic to the additive group R. We
can use the same isomorphism.
31. Suppose that φ : G → G is an isomorphism. Show that φ−1 : G → G is an
isomorphism.
Because φ−1 is also a bijective map, it suffices to show the operation preserving
property. For x, y ∈ G, there are a, b such that φ(a) = x, φ(b) = y. Then φ(ab) =
φ(a)φ(b) = xy. So
φ−1 (xy) = ab = φ−1 (x)φ−1 (y)
and φ−1 is an isomorphism.
32. Suppose that φ : G → G is an isomorphism. Show that if K is a subgroup of G,
then φ(K) = {φ(k) | k ∈ K} is a subgroup of G.
Because φ(e) ∈ φ(K), φ(K) = ∅. Let x, y ∈ φ(K). Then x = φ(a), y = φ(b)
for some a, b ∈ K. Then xy = φ(a)φ(b) = φ(ab) ∈ φ(K) because ab ∈ K. Also
because a−1 ∈ K, x−1 = (φ(a))−1 = φ(a−1 ) ∈ φ(K). By subgroup test 1, φ(K) ≤
G.
34. Prove or disprove that U (20) and U (24) are isomorphic.
In U (20), 32 = 9, 33 = 27 = 7, 34 = 81 = 1. So |3| = 4. On the other hand, in U (24),
all non-identity elements have order two. Therefore they are not isomorphic to
each other.
39. Let C be the complex numbers and
M=
a −b
b a
a, b ∈ R .
Prove that C and M are isomorphic under addition and that C∗ and M ∗ , the
nonzero elements of M , are isomorphic under multiplication.
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MATH 3005 Homework Solution
Han-Bom Moon
Define a map φ : C → M as φ(a + bi) =
φ(
a −b
b a
+
c −d
d c
a −b
b a
) = φ(
. It is bijective.
a + c −(b + d)
(b + d) (a + c)
= (a + c) + (b + d)i = (a + bi) + (c + di) = φ(
a −b
b a
)
c −d
d c
) + φ(
)
So φ is an isomorphism.
On the other hand, if we restrict φ to C∗ , then φ is a bijective map between C∗ and
M ∗ , and
φ(
a −b
b a
c −d
d c
) = φ(
ac − bd −(ad + bc)
ad + bc
ac − bd
= ac − bd + (ad + bc)i = (a + bi)(c + di) = φ(
a −b
b a
)φ(
)
c −d
c d
).
Therefore φ restricted to C∗ is an isomorphism between C∗ and M ∗ .
48. Let φ be an isomorphism from a group G to a group G and let a belong to G.
Prove that φ(C(a)) = C(φ(a)).
Let x ∈ φ(C(a)). Then there is y ∈ C(a) such that φ(y) = x. xφ(a) = φ(y)φ(a) =
φ(ya) = φ(ay) = φ(a)φ(y) = φ(a)x. So x ∈ C(φ(a)). Therefore φ(C(a)) ⊂
C(φ(a)).
Conversely, suppose that x ∈ C(φ(a)). There is y ∈ G such that φ(y) = x. φ(ya) =
φ(y)φ(a) = xφ(a) = φ(a)x = φ(a)φ(y) = φ(ay). Because φ is one-to-one, ya =
ay and y ∈ C(a). Therefore x = φ(a) ∈ φ(C(a)). So C(φ(a)) ⊂ φ(C(a)) and
C(φ(a)) = φ(C(a)).
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