Math 561 H Fall 2011
Exam 2 — Mon Oct 31 Drew Armstrong
There are 3 problems with a total of 9 sections. This is a closed book test. Any student
caught cheating will receive a score of zero. In any of the 9 sections, you may assume the
results from the other sections.
1. Consider a subgroup H ≤ G and two elements a, b ∈ G.
(a) Prove that aH = bH implies a−1 b ∈ H. (Hint: Note that b ∈ bH.)
Proof. Suppose that aH = bH. Since b ∈ bH = aH, there exists h ∈ H such that
b = ah. But then a−1 b = h ∈ H.
(b) Prove that a−1 b ∈ H implies aH = bH. (You need aH ⊆ bH and bH ⊆ aH.)
Proof. Suppose that a−1 b = h ∈ H. In order to show aH = bH we must show
aH ⊆ bH and bH ⊆ aH. So consider an arbitrary element ak ∈ aH with k ∈ H.
Then we have ak = (bh−1 )k = b(h−1 k) ∈ bH, hence aH ⊆ bH. The proof of
bH ⊆ aH is similar.
2. Let G = hgi be a cyclic group with a subgroup H ≤ G.
(a) Prove that ϕ(n) := g n defines a surjective homomorphism ϕ : Z → G.
Proof. By definition, every element of G = hgi has the form g n for some n ∈ Z,
hence the map is surjective. It is a homomorphism because ϕ(m + n) = g m+n =
g m g n = ϕ(m)ϕ(n) for all m, n ∈ Z.
(b) Prove that ϕ−1 (H) := {n ∈ Z : ϕ(n) ∈ H} is a subgroup of Z. It follows that
ϕ−1 (H) = aZ for some a ∈ Z (you don’t need to prove this).
Proof. First note that 0 ∈ ϕ−1 (H) since ϕ(0) = g 0 = 1G ∈ H. Next, suppose
that n ∈ ϕ−1 (H); i.e. ϕ(n) ∈ H. But then ϕ(−n) = ϕ(n)−1 is also in H, hence
−n ∈ ϕ−1 (H). Finally, let m, n ∈ ϕ−1 (H); i.e. ϕ(m) and ϕ(n) are in H. But then
ϕ(m + n) = ϕ(m)ϕ(n) is also in H, hence m + n ∈ ϕ−1 (H).
(c) Prove that H = hg a i and hence H is cyclic.
Proof. Since ϕ−1 (H) ≤ Z, we have ϕ−1 (H) = aZ for some a ∈ Z. Then by
definition we have ϕ(aZ) = H. That is, every element of H has the form ϕ(ak) =
g ak = (g a )k for some k ∈ Z. We conclude that H = hg a i. (In particular, H is
cyclic.)
3. Consider two finite subgroups H, K ≤ G with K E G a normal subgroup.
(a) Prove that HK := {hk : h ∈ H, k ∈ K} is a subgroup of G.
Proof. First note that 1G ∈ HK because 1G ∈ H ∩ K, hence 1G = 1G · 1G ∈ HK.
Next, consider g ∈ HK. Then there exist h ∈ H, k ∈ K such that g = hk. We wish
to show that g −1 = k −1 h−1 ∈ HK. But k −1 h−1 ∈ Kh−1 = h−1 K means there
exists k 0 ∈ K such that k −1 h−1 = h−1 k 0 ∈ HK. Finally, consider h1 k1 and h2 k2 in
HK. We wish to show that h1 k1 h2 k2 ∈ HK. Indeed, since k1 h2 ∈ Kh2 = h2 K,
there exists k 00 ∈ K such that k1 h2 = h2 k 00 . Hence h1 k1 h2 k2 = h1 h2 k 00 k2 ∈ HK.
(b) Since K E HK we can form the quotient group (HK)/K. Prove that the map
ϕ(h) := hK is a surjective homomorphism ϕ : H → (HK)/K.
Proof. The map is a homomorphism since ϕ(ab) = (ab)K = (aK)(bK) = ϕ(a)ϕ(b).
Then note that each coset in HK/K looks like (hk)K = hK for some h ∈ H,
k ∈ K. In this case we have ϕ(h) = hK = (hk)K, so the map is surjective.
(c) Prove that the kernel of ϕ is H ∩ K.
Proof. Note that ϕ(h) = hK = K if and only if h ∈ K. Hence h ∈ H is in the
kernel if and only if h is also in K. We conclude that ker ϕ = H ∩K. (In particular,
this proves that H ∩ K E H.)
(d) Use the First Isomorphism Theorem and Lagrange’s Theorem to prove that
|H| · |K|
|HK| = .
|H ∩ K|
Proof. By the First Isomorphism Theorem we have H/ ker ϕ ≈ imϕ, which by parts
(b) and (c) says that H/(H ∩ K) ≈ (HK)/K. Applying Lagrange’s Theorem to
both sides gives |H|/|H ∩ K| = |HK|/|K|. Then multiply both sides by |K|.
