Test 1 – Solutions
1. Suppose G is a group, a and b are elements of G of finite order, and ab = ba.
(a) Prove that |ab| is a divisor of lcm(|a|, |b|). Note, lcm denotes least common multiple.
Let n = |a|, m = |b|, and let r be a common multiple of n and m. Then there exist integers s and
t such that r = ns and r = mt (because r is a common multiple of n and m). Then
(ab)r = ar br
n s
because ab = ba
m t
= (a ) (b )
= es et
=e
So, r is a multiple of the order of ab. In particular, this applies to lcm(n, m).
(b) Disprove: |ab| = lcm(|a|, |b|).
Take a = b = 1 in Z2 . Then |a| = |b| = 2, so lcm(|a|, |b|) = 2, but ab = 1 + 1 = 0 in Z2 (operation
in Z2 is addition), and |ab| = |0| = 1.
(c) Prove that if |a| and |b| are relatively prime, then |ab| = |a| · |b|.
Again let n = |a| and m = |b|. Since gcd(n, m) = 1, there exist integers r and s such that
rn + sm = 1. By our proof in part (a), (ab)nm = e, so we just have to show that this is the least
power. Now if k > 0 such that (ab)k = e, then ak bk = e (since ab = ba), and so bk = a−k .
Consider hbk i = ha−k i. This is a subgroup of hbi, and a subgroup of hai. By Lagrange’s theorem,
the order of hbk i then divides n and m, which have gcd of 1, so hbk i = {e}. But then a−k = bk = e,
which implies ak = e. Thus k is a multiple of |a| and of |b|. Since |a| and |b| are relatively prime,
k is a multiple of |a| · |b|, and we are done.
2. Let G be a group and H a subset of G. Prove that H ≤ G iff H 6= ∅ and for all a, b ∈ H, ab−1 ∈ H.
(⇒) Since e ∈ H, H is non-empty. Now if a, b ∈ H, b−1 ∈ H because H is closed under inverse. Then
using that H is closed under product we get that ab−1 ∈ H.
(⇐) We will show that e ∈ H, that H is closed under inverse, and that H is closed under product.
Since H is not empty, there exists h ∈ H. Applying the condition we get e = hh−1 ∈ H. Now if h ∈ H,
h−1 = eh−1 ∈ H from the condition (we already showed e ∈ H. Finally if h1 , h2 ∈ H, h−1
2 ∈ H (we
−1
just showed H is closed under inverse), so h1 h2 = h1 (h−1
)
∈
H.
2
3. Determine how many homomorphisms there are in each case (with justification).
(a) Z2 → D5
Since the domain is cyclic of order 2, we just have determine the number of 2-torsion elements in
the range. But D5 has 5-reflections which have order 2, and the identity. The other 4 elements
are rotations which are non-trivial elements in the cyclic subgroup of order 5, so they have order
5. Thus the count is 5 + 1 = 6.
(b) S3 → Z5
Note that the orders of the domain and the range are relatively prime (6 and 5 respectively). So,
the image of any homomorphism has to have order 1 (it divides both the order of the domain and
the range). Thus, there is only one homomorphism, the trivial homomorphism.
4. For parts (a) and (b), determine the isomorphism types of the following abelian groups (i.e., to what
product of cyclic groups is each isomorphic to). Justify your answers.
(a) Z∗8
Squaring each element we get 12 = 32 = 52 = 72 = 1. So this is a group of order 4 without an
element of order 4, so it is Z2 × Z2 .
(b) Z∗16
There are 8 elements (1, 3, 5, 7, 9, 11, 13, and 15), and 152 = 72 = 92 = 1, so there are at
least 3 elements of order 2, so the group cannot be cyclic. On the other hand, 32 = 9, so not all
non-trivial elements have order 2. Of the 3 abelian groups of order 8, the only possibility left is
Z4 × Z2 .
(c) Determine, up to isomorphism, all abelian groups of order 54 . No justification needed, but each
group should be listed exactly once up to isomorphism.
The groups are Z54 , Z53 × Z5 , Z52 × Z52 , Z52 × Z5 × Z5 , and Z5 × Z5 × Z5 × Z5 .
5. (a) State Lagrange’s theorem.
(b) Prove Lagrange’s theorem. (This is a theorem from the course. You may use information from
previous sections, but anything from the Lagrange’s theorem section needed for the proof is considered part of the proof.)
See the text. The main ideas are to show that left cosets of a subgroup partition the elements of the
group, and then that all left cosets have the same number of elements as the subgroup itself.
6. Suppose H and K are finite subgroups of order n of a group G.
(a) Prove that if n is prime, then either H = K or H ∩ K = hei.
Note that H ∩K is a subgroup because it is the intersection of two subgroups, and it is a subgroup
of H. Thus the order of H ∩ K divides n which is prime. Thus, |H ∩ K| = 1 or n. If |H ∩ K| = 1,
then H ∩ K = hei. If |H ∩ K| = n, then H ∩ K = H, and so H = K (since both subgroups are of
ofder n).
(b) Prove that the conclusion to part (a) is false if we do not assume that n is prime.
Take G = Q8 , H = hii and K = hji. Then H and K both have order 4, but H ∩ K = {1, −1}
which is neither H nor is it hei.
Note, groups of order 8 are the smallest to provide counterexamples, and any of the five groups
of order 8 will do except for Z8 .
(c) Determine the number of subgroups of order 7 in S7 . Explain/justify your answer.
We start by counting the number of elements of order 7. The only cycle type in S7 which produces
elements of order 7 are 7-cycles, and the number of ways to make 7-cycles from 7 things is (7 − 1)!.
Now, each element of order 7 generates a subgroup of order 7. Each subgroup of order 7 has 6
generators (any non-identity element because it has prime order). Finally, the distinct subgroups
of order 7 intersect trivially by part (a). Thus, the subgroups of order 7 partition the elements of
order 7 into subsets of size 6. Thus, the number of subgroups of order 7 is 6!/6 = 5! = 120.
7. Suppose G is an abelian group (whose operation is written multiplicatively), and m is a positive integer.
Define φm : G → G by φm (g) = g m .
(a) Prove that φm is a homomorphism.
For all a, b ∈ G,
φm (ab) = (ab)m
= am bm
because G is abelian
= φm (a)φm (b)
So, φm is a homomorphism.
(b) Prove that if G is finite and gcd(|G|, m) = 1, then φm is an automorphism of G.
If a ∈ ker φm , e = φm (a) = am . Thus, the order of a divides m. But, the order of a divides |G|.
Since gcd(|G|, m) = 1, |a| = 1 which implies a = e. Thus, ker φm = hei and so φm is 1-1. Since it
is a function from a finite set to itself and is 1-1, it is onto. We know from part a that φm is a
homomorphism, so φm ∈ Aut(G).
(c) Prove that if n, d ∈ Z such that n = md, then for G = Zn , then ker φm = φd [Zn ]. (i.e., the kernel
of φm equals the image of φd .)
On one hand (switching to additive notation for Zn ), φd [Zn ] = {dk | k ∈ Zn } = hdi. By the
formula for the order of an element in a cyclic group,
|φd [Zn ]| = |hdi| = |d| =
n
n
= =m
gcd(d, n)
d
Interchanging the roles of d and m. Note, the number of elements of Zn mapping to each of
φm [Zn ] via φm is | ker(φm )|. Thus,
d = |φm [Zn ]| =
|Zn |
n
=
| ker(φm )|
| ker(φm )|
And so, | ker(φm )| = n/d = m. So, ker(φm ) and φd [Zn ] are both subgroups of Zn of order m.
But a cyclic group has a unique subgroup for each order dividing the order of the group, so these
two subgroups must be equal: ker(φm ) = φd [Zn ].
Note, the last bit of logic can be done slightly differently. It is easy to show that the composition
φd ◦ φm is the trivial map, and so φm [Zn ] ⊆ ker(φd ). One one has deduced that they have the
same order, then they have to be equal.