Mathematics 1214: Introduction to Group Theory
Tutorial exercise sheet 4
1. The complex numbers C form a group under addition with identity element 0, such that the
inverse of z ∈ C is −z.
Consider the five sets N, Z, Q, R, C. Which of these are subgroups of (C, +)?
Solution All except N are non-empty subsets H ⊆ C such that x, y ∈ H =⇒ x + (y −1 ) =
x + (−y) = x − y ∈ H (they are closed under subtraction) so they are subgroups of (C, +) by
Theorem 11. However, 1 ∈ N and 1−1 = −1 6∈ N, so N is not a subgroup of (C, +).
2. Find a subgroup H of (R, +) such that
√
2 ∈ H and Z ⊆ H 6= R.
√
Solution For example, let H = {n + m 2 : n, m ∈ Z}. Then Z ⊆ H and H ⊆ R, so
∅=
6 H ⊆ R and x, y ∈ H =⇒ x + y ∈ H and x ∈ H =⇒ −x ∈ H, so H is a subgroup
of (R, +).
√
√
√
We claim that H 6=√R. Indeed, suppose that
3 ∈ H.
Then 3 = n + m √2 for some integers
√
√
n, m, so 3 = (n+m 2)2 = n2 +2m2 +nm 2. Since 2 is not rational, nm 2 = 3−n2 −2m2 ∈
Z =⇒ nm = 0, so n = 0 or m = 0. If m = 0 then 3 = n2 , which is false since 3 is not the
square of an integer. So n = 0, so 3 = √
2m2 , so 3 is divisible
√ by 2, which is false again. So we
have a contradiction. This shows that 3 6∈ H, and since 3 ∈ R, this shows that H 6= R.
3. Recall that An is the set of even permutations in Sn , which is a group under composition, and
the order of this group is 21 |Sn |.
(a) Show that A3 = {(1), (1 2 3), (1 3 2)}.
(b) How many subsets of A3 are there? Which of these are subgroups?
(c) What is the order of the group A4 ? List the elements of A4 , writing them as products of
disjoint cycles. [We can write the identity permutation as (1)].
Solution (a) We have A3 = {α ∈ S3 : α is even}. Now
S3 = {(1), (1 2 3), (1 3 2), (1 2), (1 3), (2 3)}.
A cycle of length 3 is even, a cycle of length 2 is odd, and (1) is the identity permutation,
which is even. Hence A3 = {(1), (1 2 3), (1 3 2)}.
(b) We have |A3 | = 3, so there are 23 = 8 subsets. [Indeed, if we want to construct an arbitrary
subset of a set of size n, then there are two possibilities for each element: do we include it in
n
the set, or not? Hence there is a total of 2| × 2 ×
{z· · · × 2} = 2 possible subsets.]
n times
Any subgroup of A3 must contain the element (1) by Theorem 9, since (1) is the identity
element of A3 .
Only four subsets of A3 contain (1). These are:
{(1)},
S = {(1), (1 2 3)},
T = {(1), (1 3 2)},
A3 .
We know that {(1)} and A3 are subgroups of A3 . On the other hand, (1 2 3) ∈ S but
(1 2 3)(1 2 3) = (1 3 2) 6∈ S, so S is not a subgroup. Similarly, (1 3 2) ∈ T but (1 3 2)(1 3 2) =
(1 2 3) 6∈ T , so T is not a subgroup.
So the only subgroups of A3 are {(1)} and A3 .
(c) The order of A4 is |A4 | = 21 |S4 | = 21 4! = 12.
We know that 3-cycles are even permutations. The 3-cycles in S4 are:
(1 2 3),
(1 3 2),
(1 2 4),
(1 4 2) (1 3 4),
(1 4 3),
(2 3 4),
(2 4 3).
Clearly, these eight permutations are all distinct (that is, no two of them are equal).
We also know that the product of two 2-cycles (transpositions) is even. If we require the
2-cycles to be disjoint, then possibilities are:
(1 2)(3 4),
(1 3)(2 4),
(1 4)(2 3).
Clearly, no two of these are equal. Moreover, none of them fixes any element of {1, 2, 3, 4} but
each of the 3-cycles above does. So we have found 11 different elements of A4 . Together with
the identity permutation (1), this completes the list.
4. Let (G, ∗) be a group with identity element e. For x ∈ G, let us write x0 = e and xk =
x
| ∗ x ∗{z· · · ∗ x} for k ∈ N.
k times
Suppose that m, n are non-negative integers. Prove that:
(a) xm ∗ xn = xm+n .
(b) if m > n and xm = xn , then xm−n−1 = x−1
Use (b) to show that if (G, ∗) is a finite group and H is a non-empty subset of G such that
x, y ∈ H =⇒ x ∗ y ∈ H, then H is a subgroup of (G, ∗).
[Hint: if x ∈ H, show that (at least) two of the elements in list x, x2 , x3 , x4 , . . . must be equal.]
Solution (a)
n+m
xm ∗ xn = (x
.
| ∗ x ∗{z· · · ∗ x}) ∗ (x
| ∗ x ∗{z· · · ∗ x}) = x
| ∗ x ∗{z· · · ∗ x} = x
n times
m times
n + m times
(b)
xm = xn =⇒
=⇒
=⇒
=⇒
e = xn ∗ (xm )−1
xm−n = xm−n ∗ e = xm−n ∗ xn ∗ (xm )−1 = xm−n+n ∗ (xm )−1 = xm ∗ (xm )−1 = e
xm−n−1 ∗ x = xm−n−1+1 = e and x ∗ xm−n−1 = x1+m−n−1 = e
xm−n−1 = x−1 .
Now suppose that (G, ∗) is a finite group and let H be a non-empty subset of G. If x ∈ H
then X = {x, x2 , x3 , x4 , . . . } is a subset of G. If xn 6= xm for all integers n, m ≥ 0 with m > n,
then X is an infinite set, which is impossible since X ⊆ G and G is finite. So there must exist
integers m, n ≥ 0 with m > n such that xm = xn . By (b), xm−n−1 = x−1 .
If m − n − 1 = 0 then x−1 = x0 = e, so x = (x−1 )−1 = e−1 = e. So x−1 = e ∈ H.
If m − n − 1 ≥ 1 then xm−n−1 = x−1 ∈ H; indeed, since x, y ∈ H =⇒ x ∗ y ∈ H, we have
x2 = x ∗ x ∈ H and x3 = x2 ∗ x ∈ H and, in general, xk ∈ H for all k ≥ 1.
Thus x ∈ H =⇒ x−1 ∈ H, and by hypothesis, H is a non-empty subset of G such that
x, y ∈ H =⇒ x ∗ y ∈ H. Hence H is a subgroup of (G, ∗).
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5. Let (G, ∗) be a group and let H ⊆ G. Suppose that ∗H is defined by x ∗H y = x ∗ y for x, y ∈ H,
and ∗H is a well-defined operation on H such that (H, ∗H ) is a group.
Let e be the identity element of the group (G, ∗).
Let eH be the identity element of the group (H, ∗H ).
If x ∈ G, let x−1 be the inverse of x with respect to the operation ∗.
If x ∈ H, let x−1,H be the inverse of x with respect to the operation ∗H .
Prove that:
(a) H 6= ∅
(b) x, y ∈ H =⇒ x ∗ y ∈ H
(c) e = eH
[Hint: show that eH ∗ eH = eH and consider e−1
H .]
(d) x ∈ H =⇒ x−1 = x−1,H
(e) x ∈ H =⇒ x−1 ∈ H
(f) H is a subgroup of (G, ∗)
[Together with Theorem 9, this exercise shows that H is a subgroup of (G, ∗) if and only
if (H, ∗H ) is a group].
Solution (a) Since (H, ∗H ) is a group, H contains an identity element for ∗H . So H 6= ∅.
(b) ∗H is a well-defined operation on H, so ∗H : H × H → H. If x, y ∈ H then this shows that
x ∗H y ∈ H. But x ∗ y = x ∗H y, so x ∗ y ∈ H.
(c)
eH is the identity element for ∗H =⇒
=⇒
=⇒
=⇒
=⇒
=⇒
eH ∗H eH = eH
eH ∗ eH = eH
−1
e−1
H ∗ (eH ∗ eH ) = eH ∗ eH
(e−1
H ∗ eH ) ∗ eH = e
e ∗ eH = e
eH = e.
(d) Let x ∈ H. By (c), e is the identity element of the group (H, ∗H ). We have x ∗H x−1,H =
x−1,H ∗H x = e, so x ∗ x−1,H = x−1,H ∗ x = e. By the uniqueness of inverses in the group (G, ∗),
this shows that x−1 = x−1,H .
(e) Let x ∈ H. By the definition of x−1,H , we have x−1,H ∈ H, and by (d), we have x−1 = x−1,H .
Hence x−1 ∈ H.
(f) We have H ⊆ G, H 6= ∅, x, y ∈ H =⇒ x ∗ y ∈ H and x ∈ H =⇒ x−1 ∈ H. This is
precisely the definition of a subgroup of (G, ∗), so H is a subgroup of (G, ∗).
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