MAS4010 ADVANCED TOPICS IN ALGEBRA
SOLUTIONS TO EXERCISE SHEET 3
Question 1
Let G be the subgroup of S7 generated by the two permutations a = (1234567)
and b = (124)(365).
(i) Show that |G| is a multiple of 21.
(ii) Show that bab−1 = a2 , and deduce that |G| = 21.
(iii) Find the conjugacy classes of G (there are 5 of them), and find the
centraliser of one representative of each conjugacy class.
(iv) What is the centre of G?
Solution:
(i) The permutations a and b have order 7, 3 respectively, so they generate
cyclic subgroups of G of order 7 and 3. By Lagrange’s Theorem, |G|
is divisible by the order of any subgroup, so |G| is divisible by both 3
and 7. Hence |G| is divisible by 21.
[4 marks]
(ii) bab−1 = (124)(365)(1234567)(142)(356) = (1357246)
and a2 = (1234567)(1234567) = (1357246). Thus bab−1 = a2 .
[3 marks]
2
We can rewrite this as ba = a b. Any element g of G can be written as
a product of terms of the form ai or bj (in some order) with i, j ∈ Z.
Using the rule ba = a2 b, together with the facts that a7 = e = b3 , we can
move all occurrences of b to the right (possibly changing the powers of
a as we do so), and so write g in the form ai bj with i ∈ {0, 1, 2, 3, 4, 5, 6}
and j ∈ {0, 1, 2}. Thus |G| ≤ 21. Combining this with (i), we have
|G| = 21.
[2 marks]
(iii) e is obviously a conjugacy class of size 1, with centraliser the full group
G.
As bab−1 = a2 we have bak b−1 = a2k for any integer k, so that a, a2 and
a4 are all conjugate, as are a3 , a6 and a5 . Thus each of these powers of
a lies in a conjugacy class of size at least 3, and its centraliser contains
the the cyclic subgroup hai of order 7. Since the size of the conjugacy
class multiplied by the order of the centraliser is |G| = 21, it follows
that we have found everything in the conjugacy class and everything
in the centraliser. Thus two of the conjugacy classes are {a, a2 , a4 } and
{a3 , a6 , a5 }, and the centraliser of any element in either of these is hai.
From bab−1 = a2 we have a−1 ba = ab and thus a−1 (ai b)a = ai+1 b for
any i ∈ Z. Hence the 7 elements ai b, 0 ≤ i ≤ 6, are all conjugate. Each
element in this conjugacy class has order 3, and its centraliser contains
the subgroup it generates. Thus {ai b | 0 ≤ i ≤ 6} is a conjugacy
1
class, and its representative b has centraliser hbi of order 3. Similarly,
{ai b2 | 0 ≤ i ≤ 6} is a conjugacy class, and its representative b2 has
centraliser hbi.
[10 marks]
(iv) Z(G) consists of all elements in conjugacy classes of size 1. From (iii)
the only conjugacy class of size 1 in G is {e}, so Z(G) = {e}, i.e. this
group G has trivial centre.
[3 marks]
Total: 22 marks
2
Question 2
Let H be a subgroup of index 2 in a group G. Show that if N is any normal
subgroup of G then N ∩ H is a normal subgroup of H and has index either
1 or 2 in N . Given that An is simple for n ≥ 5, deduce that An the only
proper normal subgroup of Sn .
Solution:
Clearly N ∩ H is a subgroup of H. To show it is normal in H, let x ∈ N ∩ H
and h ∈ H. Then hxh−1 ∈ N since N is normal in G, and hxh−1 ∈ H since
h, x ∈ H. Thus hxh−1 ∈ N ∩ H. This shows that N ∩ H is normal in H.
[4 marks]
If N ⊆ H then N ∩ H = N so N ∩ H has index 1 in N .
If N 6⊆ H we can choose n ∈ N with n 6∈ G. Since H has index 2 in G, it
follows that G is the union of the two cosets H and nH, so that N is the
union of the two cosets N ∩ H and N ∩ nH = n(n−1 N ∩ H) = n(N ∩ H).
Thus N ∩ H has index 2 in N .
We have now shown that N ∩ H has index 1 or 2 in N .
(Alternative proof: H is normal in G since it has index 2. If N 6⊆ H then
G = HN so that G/H = N H/H ∼
= N/(N ∩ H), showing that N ∩ H has
index 2 in N .)
[4 marks]
Now take G = Sn , H = An for n ≥ 5. For any normal subgroup N of Sn ,
we know that N ∩ An is a normal subgroup of An , so N ∩ An = An or {e},
since An is simple. In the first case, An ⊆ N so N = An or Sn . In the second
case, since N ∩ An = {e} has index 1 or 2 in N , we have that |N | = 1 or 2.
Now clearly any subgroup of order 2 in Sn cannot be normal: every element
of order 2 has at least 2 conjugates if n ≥ 3. Thus we have N = {e}. Hence
the only normal subgroups of Sn are {e}, An and Sn .
[4 marks]
Total: 12 marks
3
The aim of the next two questions is to give a proof that An is
simple for n ≥ 5.
Question 3
Let n ≥ 5.
(i) Show that (ij)(kl) = (kil)(ijk) and (ij)(ik) = (ikj), where i, j, k, l
are distinct elements of {1, 2, . . . , n}.
(ii) Deduce that An is generated by 3-cycles.
(iii) Verify that (ijk) = (12i)(2jk)(12i)−1 , that (2jk) = (12j)(12k)(12j)−1
and that (1jk) = (12k)−1 (12j)(12k), where i, j, k are distinct elements
of {3, . . . , n}.
(iv) Deduce that An is generated by the 3-cycles (123), (124), . . . , (12n).
Solution:
(i) Multiplying out the permutations, we check that (kil)(ijk) = (ij)(kl)
and (ij)(ik) = (ikj).
[4 marks]
(ii) Any element of An may be written as a product of an even number of
transpositions. It can therefore be written as a product of elements,
each of which is the product of a pair of transpositions. We can assume
the two transpositions in any pair are different (otherwise their product
is the identity). So any element of An is a product of elements, each
either of the form (ij)(kl) or of the form (ij)(ik), where i, j, k, l are
distinct. Each such pair of transpositions is a product of (either one or
two) 3-cycles, by (i).
[4 marks]
(iii) Multiplying out the permutations, we check that verify that
(12i)(2jk)(12i)−1 = (ijk), that (12j)(12k)(12j)−1 = (2jk) and that
(12k)−1 (12j)(12k) = (1jk).
[4 marks]
(iv) By the 2nd and 3rd statements of (iii), each 3-cycle of the form (1jk)
or (2jk) with j, k ≥ 3 is in the subgroup of An generated by the 3cycles (123), (124), . . . , (12n). The first statement of (iii) then shows
that every 3-cycle is in this subgroup. But by (ii), An is generated by
3-cycles, so An is generated by (123), (124), . . . , (12n).
[4 marks]
Total: 16 marks
4
Question 4
Let H 6= {e} be a normal subgroup of An (with n ≥ 5) which contains
no 3-cycles. Let α 6= e be an element of H moving as few of the symbols
X = {1, 2, . . . , n} as possible. Show that, after relabelling the symbols in X
if necessary, either
(a) α = (12)(34); or
(b) α = (12)(34)(56) . . . is a product of at least three 2-cycles; or
(c) α = (123 . . .) . . . contains a cycle of length at least 3, and also moves 4
and 5.
Let γ = βαβ −1 α−1 with β = (345). Show that γ ∈ H and that in each of the
cases (a), (b) and (c), γ 6= e but γ moves fewer symbols in X than α does.
Deduce that any nontrivial normal subgroup of An contains a 3-cycle, and
hence that An is simple.
Solution:
If e 6= α ∈ H then α cannot move exactly 2 symbols (otherwise it would be
an odd permutation) or exactly 3 symbols (since H contains no 3-cycles).
Hence it must move at least 4 symbols. Suppose α moves exactly 4 symbols:
α cannot be a 4-cycle (since that would make it an odd permutation) and
the only other possibility is that α is a product of two transpositions. If α
moves more than 4 symbols, then either α has order 2, so it is a product of at
least 3 transpositions, or α contains a cycle of length at least 3 (and moves
at least 5 symbols). Hence, after renumbering the symbols if necessary, α
must have one of the forms (a), (b), (c) listed.
[2 marks]
−1 −1
−1
Now let γ = βαβ α with β = (345). We have βαβ ∈ H since H is a
normal subgroup of An , and also α−1 ∈ H, so that γ = (βαβ −1 )α−1 ∈ H.
[4 marks]
Consider the 3 cases separately:
(a) γ = (345)(12)(34)(354)(12)(34) = (354), so γ moves precisely 3 symbols.
[4 marks]
(b) γ = (345)(12)(34)(56) . . . (354)(12)(34)(56) . . .. We have γ(i) = i for all
i ≥ 6, and checking the effect of the above product on all i ≤ 6, we find
γ = (35)(46).
[4 marks]
(c) γ = (345)(123 . . .) . . . (354)(321 . . .) . . .. If α(j) = j then j > 5 and hence
β(j) = j, so that γ(j) = j. So any symbol moved by γ is also moved by α.
But γ(2) = 2 and γ(3) = 4.
[4 marks]
In each case, we have shown that γ 6= e, but that γ moves fewer symbols
that α. This contradicts the choice of α. Hence our original assumption
is invalid: H cannot be a nontrivial normal subgroup of An containing no
3-cycle. Thus any normal subgroup H 6= {e} of An contains a 3-cycle ρ. But
H then contains every 3-cycle in Sn , since every 3-cycle is conjugate in An to
either ρ or ρ−1 . As An is generated by 3-cycles (Question 3), it follows that
H = An . Thus the only normal subgroups of An are {e} and An itself; that
is, An is simple.
[2 marks]
Total: 16 marks
5
Question 5
In the group G = GL2 (Fp ) of invertible 2 × 2 matrices over the field Fp of p
elements, let g be the matrix
1 1
g=
.
0 1
Find (i) the centraliser CG (g) of g, and (ii) the normaliser NG (hgi) of the
cyclic subgroup hgi generated by g. What are the orders of these groups?
How many conjugates does g have? How many conjugates does the subgroup
hgi have?
Solution:
(i) Consider the matrix
A=
a b
c d
.
Then A is in the centraliser of g in G = GL2 (Fp ) if and only if A is invertible
(i.e. in GL2 (Fp ) ) and Ag = gA. The second condition is
a a+b
a+c b+d
=
,
c c+d
c
d
or, equivalently, c = 0, d = a. The resulting matrix will be invertible if and
only if a 6= 0. Hence
a b
×
CG (g) =
| a ∈ Fp , b ∈ Fp .
0 a
[6 marks]
Thus |CG (g)| = p(p − 1), and g has |G|/p(p − 1) = p2 − 1 conjugates.
[4 marks]
The group hgi consists of all matrices
1 m
m
g =
.
0 1
Now A is in the normaliser of hgi if A is invertible and AgA−1 = g m for some
m. Writing this as Ag = g m A, and taking A as above, we get
a = a + mc,
a + b = b + md,
c = c,
c + d = d,
which simplifies to c = 0, a = md. We need a, d 6= 0 in Fp to ensure that A
is invertible. So
a b
NG (g) =
| a, d 6= 0 .
0 d
[4 marks]
The normaliser has order p(p − 1) , so the subgroup hgi has |G|/p(p − 1)2 =
p + 1 conjugates.
[4 marks]
Total: 18 marks
2
6
Question 6
Just as over C, the characteristic polynomial of an element in GLn (Fp ) will
have the same irreducible factors as its minimal polynomial, but possibly with
higher multiplicities. When n = 2, two matrices will be conjugate if and only
if they have the same characteristic polynomial and minimal polynomial.
Using these facts, describe all the conjugacy classes of G = GL2 (Fp ), showing
how to obtain a representative of each. How many conjugacy classes are there
in GL2 (F5 )?
(You will need to consider all possibilities for the characteristic polynomial
C(X) and the minimal polynomial m(X) of a matrix in GL2 (Fp ), allowing
for the following cases: (i) C(X) has a repeated root in Fp , and m(X) has
degree 1; (ii) C(X) has a repeated root in Fp , and m(X) has degree 2; (iii)
C(X) has 2 distinct roots in Fp ; (iv) C(X) has no roots in Fp :- the last case
is analogous to a real matrix having no real eigenvalues. Remember that
roots of C(X) cannot be 0, since the matrix must be invertible.)
Solution:
(i) If m(X) has degree 1, say m(X) = X − α, then C(X) = (X − α)2 , we
can take A to be
α 0
A=
.
0 α
As there are p − 1 nonzero elements α of Fp , this gives p − 1 conjugacy
classes.
[2 marks]
(ii) If C(X) = m(X) = (X − α)2 , we again have p − 1 conjugacy classes,
one for each α 6= 0. A representative of the conjugacy class for α is
α 1
.
0 α
[2 marks]
(iii) If C(X) = m(X) = (X − α)(X − β) with α, β distinct elements of F×
p
then A is diagonalisable, and we can take
α 0
A=
.
0 β
There are (p−1)(p−2)/2 conjugacy classes of this type. (Note that the
pairs (α, β) and (β, α) correspond to conjugate matrices: so we must
count the unordered pairs (α, β).)
[3 marks]
(iv) If C(X) = m(X) and this polynomial has no roots in Fp , then C(X) =
X 2 + γX + δ where δ 6= 0 and γ 2 − 4δ is not a square in Fp . There are
p choices for γ, and then (p − 1)/2 choices for δ, provided p > 2. (For
p = 2, we see that γ 2 − 4δ = γ 2 is always a square.)
[3 marks]
For p = 5 the number of conjugacy classes is
(p − 1) + (p − 1) + (p − 1)(p − 2)/2 + p(p − 1)/2 = 4 + 4 + 6 + 10 = 24.
7
[2 marks]
Total: 12 marks
Nigel Byott
May 2004
8