Mathematics 1214: Introduction to Group Theory
Solutions to tutorial exercise sheet 9
1. Let G1 and G2 be two groups.
Prove that if H1 ≤ G1 and H2 ≤ G2 , then H1 × H2 ≤ G1 × G2 .
[Here, H1 × H2 is the set {(h1 , h2 ) : h1 ∈ H1 , h2 ∈ H2 }.]
Solution If x ∈ H1 × H2 then x = (h1 , h2 ) for some h1 ∈ H1 and some h2 ∈ H2 . Now
H1 ⊆ G1 =⇒ h1 ∈ G1 and H2 ⊆ G2 =⇒ h2 ∈ H2 . So x ∈ G1 ×G2 . Hence H1 ×H2 ⊆ G1 ×G2 .
We now have to check (SG0), (SG1) and (SG2) for H1 × H2 .
(SG0): Let e1 be the identity element of G1 and let e2 be the identity element of G2 . Since
H1 ≤ G1 we have e1 ∈ H1 , and since H2 ≤ G2 we have e2 ∈ G2 . So (e1 , e2 ) ∈ H1 × H2 . So this
set is not empty.
(SG1): If (a, b) and (c, d) are in H1 × H2 then a, c ∈ H1 ≤ G1 =⇒ ac ∈ H1 and b, d ∈ H2 ≤
G2 =⇒ bd ∈ G2 . Hence (a, b)(c, d) = (ac, bd) ∈ H1 × H2 .
(SG2): If (a, b) ∈ H1 × H2 then a ∈ H1 ≤ G1 =⇒ a−1 ∈ H1 and b ∈ H2 ≤ G2 =⇒ b−1 ∈ H2 ,
so (a−1 , b−1 ) ∈ H1 × H2 . Since this is the inverse of (a, b) [we proved this as part of showing
that G1 × G2 is a group; or you could check it again by showing that (a, b)(a−1 , b−1 ) = eG1 ×G2 ]
we have verified that (SG2) holds for H1 × H2 .
Hence H1 × H2 is a subgroup of G1 × G2 .
2. (a) Explain how the relationship Z ⊆ R allows us to identify Z × Z with a subset of the plane
P = R2 = R × R, and use this to “draw a picture” of Z × Z.
(b) Let G = Z × Z and let H = h(1, 2)i. Draw a picture of H, and draw a picture of the right
coset Ha where a = (−2, 3).
(c) The right cosets of H partition G. “Explain” this by drawing a picture. How many
(different) right cosets of H are there?
Solution (a) The picture is a bit hard to type. But you should simply plot all of the points
(n, m) where n and m are integers.
For H, we have H = h(1, 2)i = {(1, 2)k : k ∈ Z} = {(k, 2k) : k ∈ Z}. So the picture for H
should consist of these points: the points on the line y = 2x with x an integer.
(b) H(−2, 3) = {h(−2, 3) : h ∈ H} = {(k, 2k)(−2, 3) : k ∈ Z} = {(k − 2, 2k + 3) : k ∈ Z}. So
the picture should be the points with coordinates y = 2k + 3, x = k − 2 for k an integer, which
is the same as y = 2(x + 2) + 3 = 2x + 7 with x an integer. So, plot the line y = 2x + 7 and
the points on this line with integer coordinates represent H(−2, 3).
(c) A typical right coset of Ha is a translation of H by the vector a. These are parallel
integer-lines to H (where an integer-line consists of the sets of points on a line with integer
coordinates), and we obtain every integer-line parallel to H in this way.
Since they’re parallel integer-lines, no two different cosets intersect.
If a ∈ Z × Z then (0, 0)a = a is in Ha. So a is in the coset Ha. So the cosets cover Z × Z.
So they do indeed partition Z × Z.
There are infinitely many different cosets, since there are infinitely many of these parallel
integer-lines.
3. If G and H are groups with |G| > 1 and |H| > 1, prove that there are at least four different
subgroups of G × H.
[Hint: consider subgroups of the form S × T where S ≤ G and T ≤ H].
Solution Consider {eG } and G. These are different (since |G| > 1) and are subgroups of G.
Similarly, {eH } and H are different subgroups of H. By the first exercise on this sheet,
{eG } × {eH } = {(eG , eH )},
{eG } × H, G × {eH },
G×H
are subgroups of G × H, and it’s easy to check that they’re all different.
4. (a) Let G be a group. Prove that ∆ = {(g, g) : g ∈ G} is a subgroup of G × G.
(b) Find a counterexample to the following statement:
If G1 and G2 are groups, then
H ≤ G1 × G2 =⇒ H = H1 × H2 for some subgroups H1 ≤ G1 and H2 ≤ G2 .
[Hint: take G1 = G2 = Z2 and apply (a).]
Solution (a) Clearly, ∆ ⊆ G × G. Let e be the identity element of G. We have (e, e) ∈ ∆,
so ∆ 6= ∅ and (SG0) holds. If x, y ∈ ∆ then x = (g, g) and y = (h, h) for some g, h ∈ G.
So xy −1 = (g, g)(h, h)−1 = (g, g)(h−1 , h−1 ) = (gh−1 , gh−1 ) which is in ∆ since gh−1 ∈ G. By
Theorem 11, ∆ ≤ G × G.
(b) Take G1 = G2 = Z2 = {[0], [1]} under the operation ⊕ of addition modulo 2, and consider
∆ = {([0], [0]), ([1], [1])}. This is a subgroup of G1 × G2 by (a), and it has order 2. There are
only two non-empty subsets of Z2 containing the identity [0], namely {[0]} and {[0], [1]} = Z2 .
There are both subgroups, and any subgroup must contain [0]. So these are the only subgroups.
Considering all four possible products H1 × H2 for H1 and H2 each equal to either {[0]} or Z2 ,
we see that none is equal to ∆. (In fact, the only one containing both elements of ∆ is Z2 × Z2 ,
and that’s too big, since it has order 4). So
∆ ≤ G1 × G2
but ∆ 6= H1 × H2 for any subgroups H1 ≤ G1 and H2 ≤ G2
so H = ∆ is a counterexample to the statement in the question.
5. Prove that if G is a finite group of prime order, then G is abelian.
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Solution If G is a finite group of prime order, then G is cyclic by Corollary 39(b). Any cyclic
group is abelian (I think we proved this in an exercise at one point, or you could prove it now
yourself just using the definitions) so G is abelian.
Or, here’s an indirect proof that finite cyclic groups are abelian: any finite cyclic group is
isomorphic to Zn for some n ∈ N by Theorem 42, and Zn = (Zn , ⊕) is an abelian group by
Theorem 24, so G is abelian by Theorem 41(a). [This argument is a nice exercise in applying
various parts of the theory, and it would do you good to check that you understand it. But it
is something of a sledgehammer for this particular nut.]
6. A corollary to Lagrange’s theorem (Corollary 39) tells us that if n is a prime number, then
every group of order n is cyclic.
For each non-prime integer n with 4 ≤ n ≤ 14, give an example of a cyclic group of order n,
and a non-cyclic group of order n.
[Hint: the cyclic group should be easy. For the non-cyclic group, if n is even with n ≥ 6 then
you should be able to think of a non-abelian group of order n, and then consider the remaining
cases, n = 4 and n = 9 separately.]
[Remark: the case n = 15 is different. Even though 15 is not prime, it turns out that every
group of order 15 is cyclic!].
Solution Zn is a cyclic group of order n for any integer n ≥ 1.
If n is even with n ≥ 6 then n = 2k for some integer k ≥ 3, so Dk is a non-abelian, hence
non-cyclic, group of order 2k = n.
The remaining cases are:
n = 4: Z2 × Z2 is a non-cyclic group of order 4.
n = 9: Z3 × Z3 is a non-cyclic group of order 9, since |Z3 × Z3 | = |Z3 | · |Z3 | = 3 · 3 = 9 and
([k], [ℓ])3 = ([3k], [3ℓ]) = ([0], [0]) = eZ3 ×Z3 for any element ([k], [ℓ]) of Z3 × Z3 , so every element
has order at most 3. So no element has order 9. So it can’t be cyclic (if it were equal to hai,
then we would have o(a) = 9).
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