Introduction to Group Theory: Some exam style questions
on sections 15 to 18: Solutions.
1. Section A
1.(a) Define
(1) a homomorphism ϕ : G → H of groups,
(2) the kernel of a homomorphism,
(3) a normal subgroup.
Solution:
(1) A function ϕ : G → H is a homomorphism if ϕ(xy) = ϕ(x)ϕ(y)
for all x, y ∈ G.
(2) The kernel ker(ϕ) of a homomorphism ϕ : G → H is {x ∈ G : ϕ(x) = eH }
(where eH is he identity element of H.
(3) A subgroup N ≤ G is normal if gng −1 ∈ N for all n ∈ N and
g ∈ G.
(b) Give an example of a subgroup of a group that is not normal,
and explain why.
Solution: Let G = D6 , and H = hbi. Then a ∈ G and b ∈ H, but
aba−1 = a2 b 6∈ H. So H is not a normal subgroup of G.
(c) Prove that the kernel of a group homomorphism ϕ : G → H is a
normal subgroup of G.
Solution: Let g ∈ G and n ∈ ker(ϕ). Then
ϕ(gng −1 ) = ϕ(g)ϕ(n)ϕ(g)−1 = ϕ(g)eH ϕ(g)−1 = ϕ(g)ϕ(g)−1 = eH ,
so gng −1 ∈ ker(ϕ).
2.(a) Define a homomorphism of groups.
Solution: Let G and H be groups. A function ϕ : G → H is a
homomorphism if ϕ(xy) = ϕ(x)ϕ(y) for all x, y ∈ G.
(b) For each of the following properties, give an example of a homomorphism that has that property.
(1) Both injective and surjective.
(2) Injective but not surjective.
(3) Surjective but not injective.
(4) Neither injective nor surjective.
Solution:
(1) For any group G, the identity homomorphism G → G is a
bijective homomorphism. For example, G could be the trivial
group {e} and ϕ : G → G given by ϕ(e) = e.
(2) For any group H and any proper subgroup G < H, the inclusion
map G → H is an injective but not surjective homomorphism.
For example, H = D6 , G = {e} and ϕ : G → H given by
ϕ(e) = eH .
(3) For any non-trivial group G, the trivial homomorphism G →
{e} is surjective but not injective. For example, let G = D6 ,
H = {e}, and ϕ : G → H given by ϕ(x) = e for all x ∈ G.
(4) For any non-trivial groups G and H, the trivial homomorphism
G → H is neither injective nor surjective. For example, take
G = H = D6 and ϕ : G → H given by ϕ(x) = eH for all x ∈ G.
[In an exam, you’d just need to give one example for each.
I’ve also given more general answers above, as you may have
chosen different groups from me.]
3.(a) Let N be a subgroup of a group G. What does it mean to say
that N is a normal subgroup of G?
Solution: N is a normal subgroup of G if gng −1 ∈ N for all g ∈ G
and n ∈ N .
(b) Give an example of
(1) a non-trivial proper normal subgroup,
(2) a subgroup that is not normal.
Solution:
(1) The group 2Z of even integers is a normal subgroup of (Z, +).
[Or hai < D2n , or An < Sn for n > 1, or many other possibilities.]
(2) hbi < D6 . [Of course, other examples would do as well.]
(c) Let N E G be a normal subgroup. Define the quotient group
G/N (you should define the group operation, but need not prove that
it is a group).
Solution: G/H is the set
{xH : x ∈ G}
of left cosets of H in G, with group operation (written multiplicatively)
(xH)(yH) = (xy)H
for x, y ∈ G.
4. Let ϕ : G → H be a group homomorphism.
(a) Prove that the image of ϕ is a subgroup of H.
Solution:
(1) ϕ(eG ) = eH , so eH ∈ im(ϕ)
(2) If h1 , h2 ∈ im(ϕ) then h1 = ϕ(g1 ) and h2 = ϕ(g2 ) for some
g1 , g2 ∈ G. So
h1 h2 = ϕ(g1 )ϕ(g2 ) = ϕ(g1 g2 ) ∈ im(ϕ).
So im(ϕ) is closed.
(3) If h ∈ im(ϕ) then h = ϕ(g) for some g ∈ G. So
h−1 = ϕ(g)−1 = ϕ(g −1 ) ∈ im(ϕ).
So im(ϕ) is closed under taking inverses.
(b) Prove that the kernel of ϕ is a normal subgroup of G.
Solution:
(1) ϕ(eG ) = eH , so eG ∈ ker(ϕ).
(2) If g1 , g2 ∈ ker(ϕ) then
ϕ(g1 g2 ) = ϕ(g1 )ϕ(g2 ) = eH eH = eH ,
so g1 g2 ∈ ker(ϕ).
(3) If g ∈ ker(ϕ) then
ϕ(g −1 ) = ϕ(g)−1 = e−1
H = eH ,
so g −1 ∈ ker(ϕ).
(4) If n ∈ ker(ϕ) and g ∈ G, then
ϕ(gng −1 ) = ϕ(g)ϕ(n)ϕ(g)−1 = ϕ(g)eH ϕ(g)−1 = ϕ(g)ϕ(g)−1 = eH ,
so gng −1 ∈ ker(ϕ).
(c) Explain briefly why every normal subgroup of a group G is the
kernel of some homomorphism ϕ : G → H for a suitable group H.
Solution: If N E G, then N is the kernel of the homomorphism
π : G → G/N given by π(x) = xN for all x ∈ G.
5.(a) Define a group action.
Solution: An action of a group G on a set X is a map
G × X → X,
where we denote by g · x the image of (g, x), such that
(1) e · x = x for all x ∈ X, and
(2) g · (h · x) = (gh) · x for all g, h ∈ G and x ∈ X.
(b) Suppose a group G acts on a set X, and let x ∈ X.
(1) Define the orbit G · x of x.
(2) Define the stabilizer Gx of x.
(3) Prove that Gx is a subgroup of G.
Solution:
(1) G · x = {g · x : g ∈ G}.
(2) Gx = {g ∈ G : g · x = x}.
(3) e · x = x, so e ∈ Gx . If g, h ∈ Gx then
(gh) · x = g · (h · x) = g · x = x,
so gh ∈ Gx . If g ∈ Gx then
g −1 · x = g −1 · (g · x) = (g −1 g) · x = e · x = x,
so g −1 ∈ Gx . So Gx is a subgroup of G.
6.(a) State, without proof, the Homomorphism Theorem.
Solution: Let ϕ : G → H be a homomorphism. Then
G/ ker(ϕ) ∼
= im(ϕ).
(b) Let G = (R \ {0}, ×) and H ≤ G the subgroup consisting of
positive real numbers. Show that
ϕ(x) = |x|
defines a homomorphism ϕ : G → H, and find ker(ϕ).
Solution: For x ∈ R \ {0}, |x| > 0, so ϕ is a function G → H. Let
x, y ∈ R \ {0}. Then
ϕ(xy) = |xy| = |x||y| = ϕ(x)ϕ(y),
so ϕ is a homomorphism.
x ∈ ker(ϕ) ⇔ |x| = 1 ⇔ x = 1 or x = −1,
so ker(ϕ) = {1, −1}.
(c) Use the Homomorphism Theorem to show that
G/{1, −1} ∼
= H.
Solution: By the Homomorphism Theorem,
im(ϕ) ∼
= G/{1, −1}.
But ϕ : G → H is surjective, since if x ∈ H then x = |x| = ϕ(x), and
so im(ϕ) = H. So
H∼
= G/{1, −1}.