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Introduction to Group Theory
Solutions 9
Unit website: http://people.maths.bris.ac.uk/~majcr/IGT.html
Questions 1,2,3,4 to be handed in for tutorials.
1. For which values of n > 0 is the alternating group An abelian?
Justify your answer.
Solution: If n = 1 or n = 2 then An is the trivial group, and so is
abelian. If n = 3 then |An | = 3!2 = 3, and we know that any group of
prime order is cyclic, and so is abelian.
But if n > 3, then consider σ = (1, 2, 3) and τ = (2, 3, 4), which
are both even permutations (since they are cycles of odd length. Then
στ = (1, 2)(3, 4), but τ σ = (1, 3)(2, 4), so στ 6= τ σ and hence An is
non-abelian.
So An is abelian if and only if n ≤ 3.
2. For which integers d does the alternating group A8 have elements
of order d? Justify your answer.
Solution: Consider the possible cycle lengths when we write an
element of A8 as a product of disjoint cycles, remembering that to get
an even permutation we need an even number of cycles of even length,
and that the order of the element will be the lowest common multiple
of the cycle lengths, so we only need to look at one element with each
pattern of cycle lengths:
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(1, 2, 3, 4, 5, 6, 7, 8) is not in A8 .
(1, 2, 3, 4, 5, 6, 7) is in A8 and has order 7.
(1, 2, 3, 4, 5, 6)(7, 8) is in A8 and has order 6.
(1, 2, 3, 4, 5, 6) is not in A8 .
(1, 2, 3, 4, 5)(6, 7, 8) is in A8 and has order 15.
(1, 2, 3, 4, 5)(6, 7) is not in A8 .
(1, 2, 3, 4, 5) is in A8 and has order 5.
(1, 2, 3, 4)(5, 6, 7, 8) is in A8 and has order 4.
(1, 2, 3, 4)(5, 6, 7) is not in A8 .
(1, 2, 3, 4)(5, 6)(7, 8) is not in A8 .
(1, 2, 3, 4)(5, 6) is in A8 and has order 4.
(1, 2, 3, 4) is not in A8 .
(1, 2, 3)(4, 5, 6)(7, 8) is not in A8 .
(1, 2, 3)(4, 5, 6) is in A8 and has order 3.
(1, 2, 3)(4, 5)(6, 7) is in A8 and has order 6.
(1, 2, 3)(4, 5) is not in A8 .
(1, 2, 3) is in A8 and has order 3.
(1, 2)(3, 4)(5, 6)(7, 8) is in A8 and has order 2.
(1, 2)(3, 4)(5, 6) is not in A8 .
• (1, 2)(3, 4) is in A8 and has order 2.
• (1, 2) is not in A8 .
• e is in A8 and has order 1.
So the orders that occur for elements of A8 are 1, 2, 3, 4, 5, 6, 7 and
15.
3. Let G and H be groups, with G = hgi an infinite cyclic group, and
let h ∈ H. Prove that there is exactly one homomorphism ϕ : G → H
for which ϕ(g) = h.
Solution: If ϕ is a homomorphism, then ϕ(g i ) = ϕ(g)i for any
i ∈ Z, and so if ϕ(g) = h then this determines ϕ(x) for any element
x = g i ∈ hgi, and so there can be at most one homomorphism with
ϕ(g) = h.
But conversely, if we define ϕ by ϕ(g i ) = hi then this is a homomorphism, since
ϕ(g i g j ) = ϕ(g i+j ) = hi+j = hi hj = ϕ(g i )ϕ(g j )
for any i, j ∈ Z.
4. Let ϕ : G → H be a group homomorphism, and let g ∈ G.
(1) Prove that ord (ϕ(g)) divides ord(g).
(2) Deduce that if hcf(|G|, |H|) = 1 then the only homomorphism
ϕ : G → H is the trivial homomorphism.
Solution:
(1) Suppose ord(g) = d. Then ϕ(g)d = ϕ(g d ) = ϕ(eG ) = eH and so
ord (ϕ(g)) divides d (if hd = e then ord(h) divides d).
(2) Suppose hcf(|G|, |H|) = 1, let ϕ : G → H be a homomorphism,
and let g ∈ G. Then ord(g) divides |G| by Lagrange’s Theorem, and so by (1) ord (ϕ(g)) divides |G|. But by Lagrange’s
Theorem ord (ϕ(g)) also divides |H|, so ord (ϕ(g)) = 1 since
hcf(|G|, |H|) = 1, and so ϕ(g) = eH .
5. Let Cm and Cn be cyclic groups of order m and n. Find all homomorphisms ϕ : Cm → Cn .
Solution: Let x and y be generators of Cm and Cn respectively.
If ϕ : Cm → Cn is a homomorphism, then ord (ϕ(x)) divides m by
n
question 4. Since the order of y k is hcf(n,k)
, if ϕ(x) = y k then we need
n
this to divide m, and so k must be a multiple of hcf(m,n)
. Conversely, if k
n
is a multiple of hcf(m,n) then there is a homomorphism with ϕ(xi ) = y ik
(this is well-defined, since if xi = xj then m divides i − j, so mk
divides ik − jk and so n, which divides mk, divides ik − jk). So there
are hcf(m, n) homomorphisms, one with ϕ(x) = y k for each of the
n
multiples k of hcf(m,n)
between 0 and n − 1.
6. [Hard] To celebrate the beginning of spring, 100 prisoners are given
a chance to win their freedom. In a room there are 100 boxes numbered
1 to 100. Each prisoner’s name is written on a piece of paper, which
is put, at random, into one of the boxes, with a different name in each
box.
Then, one at a time, each prisoner is led into the room and can
choose a box to open, read the name in it, put the name back and
close the box. He can then repeat this for another box, and so on. If
he finds his own name in the first 50 boxes that he opens then he wins.
But all 100 prisoners must win in order for them all to be released:
if even one prisoner loses, they all stay in prison.
The prisoners cannot communicate after the first prisoner opens his
first box, but they can agree a strategy in advance. Is there a strategy
that gives them a reasonable chance of freedom?
Solution: First, for comparison, consider what happens if the prisoners all choose boxes independently and at random. Each prisoner has
a 50% chance of finding his own name in the first 50 boxes he opens,
so the probability that all 100 prisoners find their own name is only
0.5100 ≈ 7.9 × 10−31 .
They can do much better with the following strategy. Each prisoner
is assigned a different number between 1 and 100. When each prisoner
goes into the room, he first opens the box with his number. Then he
opens the box with the number of the prisoner whose name he sees,
and so on.
To see how often this works, define a permutation f of {1, 2 . . . , 100}
with f (i) equal to the number of the prisoner whose name is in box i.
Then the boxes that prisoner i opens are
i, f (i), f 2 (i), . . . .
He will find his own name in the first 50 boxes if (and only if) when the
permutation f is written as a product of disjoint cycles, his number is
in a cycle of length 50 or less. And all the prisoners will find their own
names if there is no cycle of length greater than 50.
Let’s count the number
of permutations with a cycle of length exactly
100
k > 50. There are k ways of choosing the k numbers in the long
cycle, (k − 1)! ways to put them into a k-cycle, and (100 − k)! ways to
permute the remaining numbers, giving
100!
100!(k − 1)!(100 − k)!
=
k!(100 − k)!
k
permutations. Since there are 100! permutations in total, and there
can’t be two cycles with length greater than 50, the probability of not
having a cycle of length greater than 50 is
1
1
1
1−
+
+ ··· +
≈ 0.31183,
51 52
100
so by following this strategy the prisoners have a greater than 30%
chance of being released. [Even with a much larger number N of prisoners, this strategy has a chance of success of about 1−loge (2) ≈ 0.30685.]
c
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