Mathematics 1214: Introduction to Group Theory
Solutions to tutorial exercise sheet 10
1. Which of the following groups is cyclic? As usual, you should prove that your answers are
correct.
(a) Z2 × Z2
(b) Z3 × Z2
(c) Z5 × Z10
(d) Z × Z
Solution (a) Every element of Z2 × Z2 has order at most 2. Indeed, if x ∈ Z2 × Z2 then x =
([n]2 , [m]2 ) for some integers n, m, so x2 = ([n]2 , [m]2 )([n]2 , [m]2 ) = ([2n]2 , [2m]2 ) = ([0]2 , [0]2 ) =
eZ2 ×Z2 , so o(x), being the smallest positive integer such that xk = e, is no greater than 2. So
for every x ∈ Z2 × Z2 we have |hxi| = o(x) ≤ 2 < 4 = |Z2 × Z2 for every x ∈ Z2 × Z2 , so
hxi =
6 Z2 × Z2 . Hence this group is not cyclic.
(b) By direct computation, we have Z3 × Z2 = h([1]3 , [1]2 )i. So this group is cyclic.
(c) Every element of Z5 ×Z10 has order at most 10. Indeed, if x ∈ Z5 ×Z10 then x = ([n]5 , [m]10 )
for some integers n, m, so x10 = ([n]5 , [m]10 )([n]5 , [m]10 ) = ([10n]5 , [10m]10 ) = ([0]5 , [0]10 ) =
eZ5 ×Z10 , so o(x), being the smallest positive integer such that xk = e, is no greater than 10. So
for every x ∈ Z5 × Z10 we have |hxi| = o(x) ≤ 10 < 50 = |Z5 × Z10 for every x ∈ Z5 × Z10 , so
hxi =
6 Z5 × Z10 . Hence this group is not cyclic.
(d) This group is not cyclic. Indeed suppose for a contradiction that it is a cyclic group. Then
there is an element x ∈ Z × Z with Z × Z = hxi. So x = (n, m) for some integers n, m ∈ Z,
and so
Z × Z = hxi = {xk : k ∈ Z} = {(kn, km) : k ∈ Z}.
If m = 0 then (0, 1) is not in this set, which is a contradiction. So we have m 6= 0.
If n = 0 then (1, 0) is not in this set, which is a contradiction. So we have n 6= 0.
Consider the element (n, −m) ∈ Z × Z. There is an integer k ∈ Z with (kn, km) = (n, −m),
and since n, m 6= 0 this gives k = 1 and k = −1, which is a contradiction.
So Z × Z cannot be cyclic.
Alternative method: draw a picture of Z × Z and h(n, m)i for a typical element (n, m) ∈ Z × Z
and show that h(n, m)i is contained in the straight line mx = ny, so can’t cover all of Z × Z
(since there’s no single straight line containing all of the points in the plane with integer
coordinates).
2. Let H be a subgroup of a group G, let a be a fixed element of G and consider the set
aHa−1 = {aha−1 : h ∈ H}.
(a) Prove that θ : H → G, h 7→ aha−1 is an injective homomorphism.
(b) What can you conclude about the relationship between H and aHa−1 ?
(c) Show that depending on the choice of G, H and a, we may have aHa−1 = H or we may
have aHa−1 6= H.
Solution (a) We have θ(h)θ(h′ ) = aha−1 ah′ a−1 = ahh′ a−1 = θ(hh′ ), for any h, h′ ∈ H. So θ
is a homomorphism H → G. Moreover, h ∈ ker θ ⇐⇒ θ(h) = e ⇐⇒ aha−1 = e ⇐⇒ h =
a−1 ea ⇐⇒ h = e, so ker θ = {e} and so θ is injective by Theorem 45(c).
(b) Hence H ≈ θ(H) = aHa−1 , by Theorem 46.
(c) If G is any group and H = {e}, then H ≤ G and aHa−1 = {aea−1 } = {e} = H.
On the other hand if G = S3 and H = h(1 2)i = {(1), (1 2)} and a = (1 3), then (1 3)(1 2)(1 3)−1 =
(1 3)(1 2)(1 3) = (2 3) ∈ aHa−1 but this is not in H. So aHa−1 6= H.
3. Prove that if G, H are any groups, then the mapping θ : G → H,
is a homomorphism, and compute its kernel and image.
θ(a) = eH for every a ∈ G
Solution If a, b ∈ G then θ(ab) = eH and θ(a)θ(b) = eH eH = eH , so θ(ab) = θ(a)θ(b). So
θ is a homomorphism. We have ker θ = {a ∈ G : θ(a) = eH } = G and the image of θ is
θ(G) = {θ(a) : a ∈ G} = {eH }.
4. Let G and H be two groups, let θ : G → H be a homomorphism and consider the group θ(G).
(a) Prove that if G is an abelian group, then so is θ(G).
(b) Prove that if G is a finite group, then so is θ(G). Give an example with |θ(G)| = |G| and
an example with |θ(G)| < |G|.
(c) Disprove the statement: “if G is an infinite group, then so is θ(G)”.
Solution (a) Let x, y ∈ θ(G). Then x = θ(a) and y = θ(b) for some a, b ∈ G. Since G is
abelian, we have ab = ba. So xy = θ(a)θ(b) = θ(ab) = θ(ba) = θ(b)θ(a) = yx. Hence xy = yx
for any x, y ∈ θ(G), so θ(G) is abelian.
(b) G is finite, so G = {x1 , . . . , xn } where n ∈ N is n = |G|. Hence θ(G) = {θ(x1 ), . . . , θ(xn )}
is a set of size at most n, so in particular, it’s a finite set. Since θ(G) is a subgroup of H, it’s
a group. So it’s a finite group.
For example, if G is any group of order 2 or more and θ : G → H, x 7→ eH for all x ∈ G, then
θ is a homomorphism and θ(G) = {eH } so |θ(G)| = 1 < |G|.
(c) Let G be any infinite group, let H be any group and again take θ : G → H, x 7→ eH for all
x ∈ G. Then, again, |θ(G)| = 1 so θ(G) is finite, despite G being infinite.
5. (a) Let θ : G → H be a homomorphism. Show that if J is a subgroup of H, then
K = {g ∈ G : θ(g) ∈ J}
is a subgroup of G.
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(b) Apply (a) with the following data:
G = GL(2, R),
H = (R× , ·),
J = (Q ∩ (0, ∞), ·),
θ = det
to deduce that the group of 2 × 2 matrices A with real entries such that det A is positive
and rational forms a group under matrix multiplication. [Start by explaining why J is a
subgroup of H.]
Solution (a) We have K ⊆ G, and since J ≤ H we have eH ∈ J and so θ(eG ) = eH ∈ J, so
eG ∈ K. So ∅ 6= K ⊆ G. If a, b ∈ K then θ(a) ∈ J and θ(b) ∈ J, so θ(b−1 ) = θ(b)−1 ∈ J, so
θ(ab−1 ) = θ(a)θ(b−1 ) ∈ J, so ab−1 ∈ K. So by Theorem 11, K is a subgroup of G.
(b) Recall that det is a homomorphism from GL(2, R) to R× . And J = (0, ∞) is a subgroup
of R× , since it’s a non-empty subset and the product and (multiplicative) inverse of positive
real numbers are both positive real numbers. Hence K = {g ∈ GL(2, R) : det g ∈ J} = {A ∈
GL(2, R) : det A > 0} is a subgroup of GL(2, R), by (a). And this is precisely the set of 2 × 2
matrices with positive determinant.
6. Let G be a group. In Exercise 3 of homework exercise sheet 2, we proved that for each x ∈ G,
the mapping Lx : G → G, y 7→ xy, is a bijection, i.e. Lx ∈ Sym(G).
Consider the mapping θ : G → Sym(G), x 7→ Lx .
(a) Prove that Lx ◦ Ly = Lxy for every x, y ∈ G.
(b) Deduce that θ is an homomorphism G → (Sym(G), ◦).
(c) Compute ker θ and deduce Cayley’s theorem: every group G is isomorphic to a permutation group [that is, a subgroup of (Sym(S), ◦), for some set S].
Solution (a) For every a ∈ G, we have (Lx ◦ Ly )(a) = Lx (Ly (a)) = Lx (ya) = xya = Lxy (a).
So Lx ◦ Ly = Lxy .
(b) So θ(x) ◦ θ(y) = Lx ◦ Ly = Lxy = θ(xy). So θ is a homomorphism.
(c) If x ∈ G then x ∈ ker θ ⇐⇒ Lx = eSym(G) = ιG =⇒ Lx (eG ) = eG ⇐⇒ xeG = eG ⇐⇒
x = eG . So ker θ = {eG }.
By Theorem 45(c), θ is an injective homomorphism, so by Theorem 46, G ≈ θ(G), which is a
subgroup of (Sym(G), ◦) by Theorem 45(a).
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