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MATH 113 - HOMEWORK 3
SOLUTIONS
6.18. gcd(42, 30) = 6, so h30i has 42/6 = 7 elements. In fact, h30i = h6i = {0, 6, 12, 18, 24, 30, 36}.
6.48. We will show that every infinite group has infinitely many subgroups. Let G be an infinite group,
and consider the collection of all its cyclic subgroups: C = {hai | a ∈ G}.
Case 1: There are infinitely many distinct subgroups in C. Then G has infinitely many subgroups, and
we are done.
Case 2: There S
are only finitely many distinct subgroups in C, so C = {H1 , H2 , . . . , Hn }. Now for all
n
a ∈ G, a ∈ hai, so i=1 Hi = G. If all of the Hi were finite, G would also be finite. Hence some Hi is infinite.
But then Hi = hai (for some a) is an infinite cyclic subgroup of G, and infinite cyclic subgroups themselves
have infinitely many subgroups. That is, the subgroups han i for n ∈ Z+ are all distinct cyclic subgroups
of H and so also of G. This contradicts our assumption that there are only finitely many distinct cyclic
subgroups of G.
6.50. We have a ∈ G generating a cyclic subgroup of order 2 (so a 6= e but a2 = e), and a is the unique
such element of G.
Let x ∈ G and consider (xax−1 )2 = xax−1 xax−1 = xaax−1 = xx−1 = e. Now there are only two elements
of G which square to the identity: e (the unique element of order 1) and a (the unique element of order 2).
Case 1: xax−1 = e. Then xax−1 x = ex, so xa = x, and a = x−1 xa = x−1 x = e, contradiction. This case
is impossible.
Case 2: xax−1 = a. Then xax−1 x = ax, so xa = ax, as was to be shown.
6.55. Zp is cyclic of order p, and we know that a is a generator of Zp for all a relatively prime to p. And
gcd(a, p) = 1 for all 0 < a < p, so every element of Zp other than 0 is a generator of Zp .
Now let H ≤ Zp be a nontrivial subgroup. Then there is some a ∈ H, a 6= 0. So H contains a generator
of Zp , and H = Zp , so H is not a proper subgroup.
We’ve shown that Zp has no nontrivial proper subgroups.
7.4. {0, 6, 12, 18, 24, 30} = h6i
7.6. {. . . , −6, −3, 0, 3, 6, . . . } = 3Z
7.14. No - the subset {1, 3, 5} is not even a subgroup, since it does not contain the identity.
7.15. In this diagram, solid arrows represent adding 1, dashed arrows represent adding 2.
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@ 0O
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5 ^i
2
3
4 ou
7.16. In this diagram, solid arrows represent adding 2, dashed arrows represent adding 5.
1
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/1
I
7O i
6 _u
52
5o
) 3

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7.18. This diagram is Z4 with generator 1 (represented by a solid arrow).
0O
/1
3o
2
This digram is V with generators a (represented by a solid arrow) and b (represented by a dashed arrow.
e
a
b
c
(1) Recall that I gave two definitions of isomorphism in class:
• A homomorphism φ : G → H is an isomorphism if it has an inverse φ−1 : H → H and φ−1 is
also a homomorphism.
• An isomorphism φ : G → H is a bijective (one-to-one and onto) group homomorphism.
We would like to show that these two definitions are equivalent.
(a) Let A and B be sets. Show that a function f : A → B has an inverse f −1 : B → A if and only
if f is bijective.
(b) Let G and H be groups. Show that a homomorphism φ : G → H has an inverse φ−1 : H → G
which is also a homomorphism if and only if f is bijective.
Solution:
(a) Suppose f has an inverse. Let b ∈ B. Then f −1 (b) ∈ A, and f (f −1 (b)) = b, so f is onto. Let
a, a0 ∈ A such that f (a) = f (a0 ). Then a = f −1 (f (a)) = f −1 (f (a0 )) = a0 , so f is one-to-one.
Now suppose f is one-to-one and onto. Then for all b ∈ B, there exists a unique a ∈ A with
f (a) = b. Existence is because f is onto, and for uniqueness, suppose we have a, a0 ∈ A with
f (a) = f (a0 ) = b. Then since f is one-to-one, a = a0 . So we can define a function f −1 : B → A
by f −1 (b) = a if and only if f (a) = b.
And indeed f −1 is the inverse of f : for all a ∈ A, if f (a) = b, then f −1 (f (a)) = f −1 (b) = a,
and for all b ∈ B, if f −1 (b) = a, then f (f −1 (b)) = f (a) = b.
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(b) Suppose φ has an inverse which is also a homomorphism. Then by part (a), φ must be bijective,
just because it has an inverse.
Conversely, suppose φ is bijective. Then by part (a), φ has an inverse, φ−1 : H → G. We
only need to show that φ−1 is also a homomorphism. Let a, b ∈ H. Taking a0 = φ−1 (a) and
b0 = φ−1 (b), we have φ(a0 ) = a and φ(b0 ) = b. Now φ−1 (ab) = φ−1 (φ(a0 )φ(b0 )) = φ−1 (φ(a0 b0 )) =
a0 b0 = φ−1 (a)φ−1 (b).
(2) Let G and H be groups, and let S be a set of generators for G. Suppose that f : G → H and
g : G → H are both homomorphisms and that ∀s ∈ S, f (s) = g(s). Show that f = g.
What we’ve shown is that a homomorphism G → H is determined by where it sends the generators
of G.
Solution: One should first show that homomorphisms preserve inverses and identity. Let φ :
G → H be a homomorphism, let eG be the identity of G, and let eH be the identity of H.
I claim that φ(eG ) = eH . Let e0 = φ(eG ). Then e0 e0 = φ(eG )φ(eG ) = φ(eG eG ) = φ(eG ) = e0 , so
canceling an e0 from both sides, we see that φ(eG ) = e0 = eH .
Now let a ∈ G. I claim that φ(a)−1 = φ(a−1 ). Indeed, φ(a)φ(a−1 ) = φ(aa−1 ) = φ(eG ) = eH .
Now we’ll show that f and g agree on all elements of G. Let a ∈ G. Since G = hSi, we can write
a = s1 . . . sn , where each si is an element of S or eG or the inverse of an element of s. I claim that f
and g agree on all these components si . If si ∈ S, then f (s) = g(s) by assumption. If si = eG , then
f (si ) = g(si ) = eH . And if si = s−1 for s ∈ S, then f (si ) = f (s−1 ) = f (s)−1 = g(s)−1 = g(s−1 ) =
g(si ).
So f (a) = f (s1 . . . sn ) = f (s1 ) . . . f (sn ) = g(s1 ) . . . g(sn ) = g(s1 . . . sn ) = g(a). Hence f = g.
(3) For groups G and H, let Hom(G, H) be the set of homomorphisms from G to H. Show that
Hom(Z12 , Z12 ) has exactly 12 elements. Show that exactly 4 of these homomorphisms are isomorphisms.
Solution: Since Z12 = h1i, a homomorphism φ : Z12 → Z12 is determined by the value of φ(1),
by Exercise 2. Since there are 12 possible values of φ(1), there are at most 12 homomorphisms.
We’ll show that each of the possibilities gives a homomorphism. Let φn be the homomorphism
determined by 1 7→ n. Then we should have φn (a) = φn (a · 1) = a · φn (1) = a · n = an.
Now we have specified what φn does to the class a ∈ Z12 by doing a computation with the number
a which is a representative for the class. Whenever we do this, it is important to check that the
function is well-defined, by checking that if b = a, then φn (b) = φn (a). Note that if b = a, then
b = a + c · 12 for some c. Then φn (b) = bn = (a + c · 12)n = an + cn · 12 = an.
This well-definedness check is easy to overlook, but it is very important. For example, if you try to
define a homomorphism Z12 → Z5 by a 7→ a · 2, you will find that this is not well-defined: although
1 = 13 in Z12 , we have 2 6= 26 in Z5 .
Finally, φn (a +12 b) = φn (a + b) = (a + b)n = an + bn = an +12 bn = φn (a) +12 φn (b), so φn is a
homomorphism.
Note that φ1 is the identity map Z12 → Z12 , so this is obviously an isomorphism. Which others
are? For any a ∈ Z12 , φn (a) = an = a · n, so every element in the range of φn is a multiple of n. If φn
is to be onto, we must be able to realize every element of Z12 as a multiple of n, that is, hni = Z12 .
In fact, if n is a generator of Z12 , then φn will be an isomorphism. φn will be onto, since it hits
every multiple of n by φn (a) = a · n, and φn must be one-to-one, since an onto map between finite
sets of the same size must also be one-to-one.
So the isomorphisms Z12 → Z12 correspond to the generators of Z12 : φ1 , φ5 , φ7 , φ11 .
(4) For a group G, an isomorphism φ : G → G is called an automorphism of G. Let Aut(G) = {φ ∈
Hom(G, G) | φ is an isomorphism} be the set of automorphisms of G.
(a) Show that (Aut(G), ◦) is a group, where ◦ is function composition.
(b) In Exercise 3 above, you showed that Aut(Z12 ) has 4 elements. Name these four elements, and
make a multiplication table for them.
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(c) In class, I asserted (without proof) that every group of order 4 is isomorphic to Z4 or the Klein
4-group V . Which of these two groups is Aut(Z12 ) isomorphic to?
Solution:
(a) Associativity: Function composition is always associative. f ◦ (g ◦ h) = (f ◦ g) ◦ h.
Identity: The identity function idG : G → G is an isomorphism (Proof: It is a homomorphism
since idG (ab) = ab = idG (a)idG (b), and it has an inverse, since idG ◦ idG = idG ), and we have
f ◦ idG = idG ◦ f = f for all f ∈ Aut(G).
Inverses: Let f ∈ Aut(G). Then by problem (1), f has an inverse f −1 : G → G which is also a
homomorphism. Indeed, f −1 is an isomorphism, since it has an inverse, f . So f −1 ∈ Aut(G).
(b) We showed Aut(Z12 ) = {φ1 , φ5 , φ7 , φ11 }. As an example of how to multiply elements of this
group, we have (φ5 ◦ φ7 )(1) = φ5 (7) = 7 · 5 = 11, so φ5 ◦ φ7 = φ11 .
◦
φ1 φ5 φ7 φ11
φ1
φ1 φ5 φ7 φ11
φ5
φ5 φ1 φ11 φ7
φ7
φ7 φ11 φ1 φ5
φ11 φ11 φ7 φ5 φ1
(c) By inspection of the multiplication table, Aut(G) ∼
=V.
(5) Show that (Q, +) is not finitely generated.
Solution: Suppose for contradiction that S = {s1 , . . . , sn } ⊂ Q is a generating set. Write all
of the fractions si over a common denominator, b, so S = { ab1 , . . . , abn }. Note also that 0 = 0b and
i
−si = −a
b .
Now let q ∈ Q. Since S generates Q, we can write q = t1 + · · · + tm , where for all i, ti ∈ S or
ti = 0 or ti = −sj for some sj ∈ S. That is, we can write q as a sum of fractions with common
denominator b. So we can simplify q by adding the numerators to a fraction with denominator b.
Then writing q in lowest terms, the denominator of q is at most b.
But there are rational numbers which cannot be written in lowest terms with denominator at
1
most b. For example, b+1
cannot be reduced. This is a contradiction.
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