Mathematics 1214: Introduction to Group Theory
Solutions to exercise sheet 11
1. Let n ≥ 3 and let Dn be the dihedral group of order 2n. Writing ρ = ρ2π/n , show that hρi is a
normal subgroup of Dn . [Hint: if r is reflection in the x-axis, then r ◦ ρ = ρ−1 ◦ r, and the reflections
in Dn are rj = ρj ◦ r for 0 ≤ j < n.] What is the order of the quotient group Dn /hρi?
Solution Let θ : Dn → {−1, 1} be given by defining θ(α) = 1 if α is a proper motion, and
θ(α) = −1 if α is an improper motion. By homework exercise sheet 5, question 5(d), θ is then a
homomorphism into the group {−1, 1} with multiplication. Moreover, the proper motions in Dn are
precisely e, ρ, ρ2 , . . . , ρn−1 (since these are orientation-preserving elements of Dn , and the remaining
elements are all reflections, hence orientation-reversing), so ker θ = {α ∈ Dn : θ(α) = 1} = hρi,
hence hρi is a normal subgroup of Dn by Theorem 47.
Alternatively, we could avoid having to dream up a suitable homomorphism, and instead use the
definition of “normal subgroup” directly. Let N = hρi. We have to check that if n ∈ N and g ∈ Dn
then gng −1 ∈ N .
If g ∈ N then gng −1 ∈ N , since N is a subgroup.
If g 6∈ N then g is one of the reflections in Dn , say g = rj = ρj ◦ r. Then g −1 = (ρj ◦ r)−1 =
r−1 ◦(ρj )−1 = r◦ρ−j . Moreover, n ∈ N = hρi, so n = ρk for some k ∈ Z. So gng −1 = ρj ◦r◦ρk ◦r◦ρ−j .
Since r ◦ ρ = ρ−1 ◦ r, we have r ◦ ρ2 = r ◦ ρ ◦ ρ = ρ−1 ◦ r ◦ ρ = ρ−1 ◦ ρ−1 ◦ r = ρ−2 ◦ r, and similarly,
we have r ◦ ρk = ρ−k ◦ r. Hence gng −1 = ρj ◦ ρ−k ◦ r ◦ r ◦ ρ−j = ρj ◦ ρ−k ◦ ρ−j = ρj−k−j = ρ−k , which
is in N = hρi. So gng −1 ∈ N .
Hence N is normal in Dn .
Finally, we have |Dn | = 2n and |hρi| = o(ρ) = n, so |Dn /hρi| =
|Dn |
|hρi|
=
2n
n
= 2.
2. Give an example of a normal subgroup of GL(2, R), and give an example of a subgroup of GL(2, R)
which is not normal.
Solution Trivial examples of normal subgroups are GL(2, R) itself, and {eGL(2,R) } = {I}. A less
trivial example is SL(2, R), which is the kernel of the determinant and so a normal subgroup (since
the determinant is a homomorphism).
1 x
check
but taking
Consider
This is a subgroup−1of GL(2,
1 0 R)
1R}.
[you should
0 1this]
1 1 H = { 0 1 : x ∈
−1
0
X = 0 1 ∈ H and A = 1 1 ∈ GL(2, R) gives A = −1 1 and AXA = −1 2 6∈ H. So H is
not normal in GL(2, R).
3. Let G be a group and let N be a normal subgroup of G.
(a) Prove that the mapping η : G → G/N defined by η(a) = N a for a ∈ G is a homomorphism.
What are the kernel and image of η?
(b) If H is another group and θ : G → H is a homomorphism with kernel N , and ϕ : G/N → H
is defined by ϕ(N a) = θ(a), then ϕ is a well-defined homomorphism (see the proof of the
Fundamental Homomorphism Theorem). Prove that θ = ϕ ◦ η.
Solution (a) For all a, b ∈ G we have η(ab) = (N a)(N b) = N (ab) = η(ab), so η is a homomorphism. The image of η is η(G) = {η(a) : a ∈ G} = {N a : a ∈ G} = G/N and the kernel is
ker η = {a ∈ G : N a = eG/N = N } = {a ∈ G : a ∈ N } = N since N a = N ⇐⇒ a ∈ N (if you like,
by Theorem 35 applied with H = N and b = e).
(b) For any a ∈ G we have (ϕ ◦ η)(a) = ϕ(η(a)) = ϕ(N a) = θ(a) by the definitions of the mappings
η and ϕ. Since ϕ ◦ η has the same domain and codomain as θ and we’ve just seen that they take
the same values at each point of their domains, this shows that ϕ ◦ η = θ.
4. Let θ : R → C× , θ(x) = eix . Prove that θ is a homomorphism with kernel h2πi, and that R/h2πi ≈
{z ∈ C× : |z| = 1}.
Solution For any x, y ∈ R we have θ(x+y) = ei(x+y) = eix eiy = θ(x)·θ(y). So θ is a homomorphism
from (R, +) to (C× , ·). Its kernel is
ker θ = {x ∈ R : θ(x) = eC× = 1} = {x ∈ R : eix = 1} = {n · 2π : n ∈ Z} = h2πi.
By the fundamental homomorphism theorem (actually, this is Corollary 50), we have
R/h2πi ≈ θ(R) = {eix : x ∈ R} = {z ∈ C× : |z| = 1}.
5. Let G be a group with identity element e.
(a) Show that {e} ⊳ G and G/{e} ≈ G. (b) Show that G ⊳ G and G/G ≈ {e}.
Solution (a) Let θ : G → G be given by θ(a) = a for all a ∈ G (so that θ is just the identity
mapping on G). Then θ is a surjective homomorphism with kernel {e}, so {e} ⊳ G by Theorem 47,
and G/{e} ≈ G by the fundamental homomorphism theorem.
(b) Let θ : G → {e} be given by θ(a) = e for all a ∈ G. Then θ is a surjective homomorphism with
kernel G, so G ⊳ G by Theorem 47 and G/G ≈ {e} by the fundamental homomorphism theorem.
6. Let G be a group.
(a) Show that if N is a normal subgroup of G, then G/N is an abelian group if and only if
aba−1 b−1 ∈ N for all a, b ∈ G.
(b) Give an example of a non-abelian group G and a normal subgroup N of G such that G/N is a
finite abelian group. [You can always do this easily with N = G. So if you want a little more
of a challenge, try to do it with N 6= G.]
(c) Give another example of a non-abelian group G and a normal subgroup N of G such that G/N
is an infinite abelian group.
Solution (a) If a, b ∈ G then we have
(N a)(N b) = (N b)(N a) ⇐⇒ N (ab) = N (ba) ⇐⇒ (ab)(ba)−1 ∈ N ⇐⇒ aba−1 b−1 ∈ N
where we have used Theorem 35 in the middle. So G/N is abelian ⇐⇒ (N a)(N b) = (N b)(N a)
for all a, b ∈ G ⇐⇒ aba−1 b−1 ∈ N for all a, b ∈ G.
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(b) For a trivial example, take N = G = S3 ; then G is non-abelian and G/N = G/G ≈ {e} and
{e} is finite and abelian, so G is finite and abelian.
For a less trivial example, take G = S3 × Z2 and let N = S3 × {[0]}. Then G is non-abelian since
S3 is non-abelian. Moreover, N is a normal subgroup of G and G/N ≈ Z2 . [You should justify the
three statements in the last sentence by considering the map π : S3 × Z2 → Z2 , π((α, [k])) = [k] for
(α, [k]) ∈ S3 × Z2 ]. Since Z2 is a finite and abelian, this shows that G/N is finite and abelian.
(c) Let G = S3 × Z and let N = S3 × {[0]}. Then (much as in (b)) G is non-abelian, N ⊳ G and
G/N ≈ Z is an infinite abelian group.
7. If G is a cyclic group and N is a subgroup of G, explain why N is a normal subgroup of G and
prove that G/N is a cyclic group. [Hint: for the second part, if G = hai, prove that G/N = hN ai.]
Solution If G is cyclic then G is abelian. Any subgroup of an abelian group is normal. So any
subgroup N of G is normal.
Since G is cyclic, we have G = hai = {ak : k ∈ Z} for some a ∈ G. So
G/N = {N x : x ∈ G} = {N (ak ) : k ∈ Z} = {(N a)k : k ∈ Z} = hN ai.
(if you like, you can justify the third equality by applying Exercise 3(a) with Theorem 40(c).)
Hence G/N is cyclic.
8. Let G be a group and let N ⊳ G.
(a) If G is a finite group, prove that G/N is a finite group.
(b) If G is an infinite group and N is a finite normal subgroup, prove that G/N is an infinite
group.
(c) Show (by giving examples) that if G is an infinite group and N is an infinite normal subgroup,
then G/N may be finite, or it may be infinite.
Solution (a) G/N is a partition of G, so if G is finite then so is G/N .
Alternatively, if n = |G| then we may write G = {a1 , . . . , an } and so G/N = {N a : a ∈ G} =
{N a1 , . . . , N an } is a set of size at most n. [Note that G/N will usually have size much smaller
than n, since for many values of i, j we will have N ai = N aj ].
Alternatively, simply recall that if G is finite then |G/N | =
so G/N is a finite group.
|G|
.
|N |
Since |N | ≥ 1, we have |G/N | ≤ |G|,
(b) Again, G/N is a partition of G. Since G is an infinite set and each coset N a in G/N has the
same size as N , which is finite, there must be infinitely many sets in this partition. So G/N is
infinite.
(c) If G is any infinite group, then G/G is finite (in fact, it has order 1).
If G = Z × Z and N = Z × {0} then N ⊳ G and G/N ≈ Z is an infinite group.
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