8. Also since
G
is a subgroup of Sym(
X
), we
234
CHAPTER 4
Abstract Algebra
see that

G

24.
But there are plenty of permutations of the vertices
1, 2, 3, and 4 which are not graph automorphisms (find one!), and so

G

<
24.
On the other hand, note that the powers
e, σ, σ
2
, σ
3
give four au
tomorphisms of the graph, and the elements
τ, στ, σ
2
τ, σ
3
τ
give four
more.
Furthermore,
since
τστ
=
σ
3
we can show that the set
{
e, σ, σ
2
, σ
3
, τ, στ, σ
2
τ, σ
3
τ
}
is closed under multiplication and hence
is a subgroup of
G
. Therefore 8

G

and so it follows that

G

= 8 and
the above set is all of
G
:
G
=
{
e, σ, σ
2
, σ
3
, τ, στ, σ
2
τ, σ
3
τ
}
.
Below is the multiplication table for
G
(you can fill in the missing
elements).
Notice that
G
has quite a few subgroups—you should be
able to find them all (Exercise 3).
◦
e
σ
σ
2
σ
3
τ
στ
σ
2
τ
σ
3
τ
e
e
σ
σ
2
σ
3
τ
στ
σ
2
τ
σ
3
τ
σ
σ
σ
2
σ
3
e
σ
2
σ
2
σ
3
e
σ
σ
3
σ
3
e
σ
σ
3
τ
τ
στ
στ
σ
2
τ
σ
3
τ
Exercises
1. Use the corollary on page 233 to give another proof of Fermat’s
Little Theorem; see page 86.
2. Suppose that
G
is a finite group of prime order
p
. Prove that
G
must be cyclic.
3. Refer to the multiplication table above for the group
G
of symme
tries of the square and list all of the subgroups.
SECTION 4.2
Basics of Group Theory
235
4. Let
G
be a group, let
H
be a subgroup, and recall the equivalence
relation ( mod
H
) defined by
g
≡
g
0
( mod
H
)
⇔
g

1
g
0
∈
H.
The equivalence classes in
G
relative to this equivalence relation
are called the (left)
cosets
of
H
in
G
.
Are the cosets also sub
groups of
G
? Why or why not?
5. Let
G
be the group of Exercise 3 and let
K
be the cyclic subgroup
generated by
στ
. Compute the left cosets of
K
in
G
.
6. Let
G
be the group of Exercise 3 and let
L
be the subgroup
{
e, τ, σ
2
, σ
2
τ
}
. Compute the left cosets of
L
in
G
.
7. Here we shall give yet another proof of the infinitude of primes.
Define, for each prime
p
the corresponding
Mersenne number
by
setting
M
p
= 2
p

1 (these are often primes themselves). Assume
by contradiction that there are only finitely many primes and let
p
be the
largest
prime. Let
q
be a prime divisor of
M
p
= 2
p

1.
Then we have, in the multiplicative group
Z
*
q
of nonzero integers
modulo
q
, that 2
p
≡
1(mod
q
). This says, by exercise 2 on page 227
that
p
is the order of 2 in the group
Z
*
q
. Apply Lagrange’s theorem
to obtain
p

(
q

1), proving in particular that
q
is a larger prime
than
p
, a contradiction.
4.2.8
Homomorphisms and isomorphisms
What is the difference between the additive group (
Z
6
,
+) and the mul
tiplicative group (
Z
*
7
,
·
)?
After all, they are both cyclic: (
Z
6
,
+) has
generator 1 (actially, [1]), and (
Z
*
7
,
·
) has generater 3 ([3]). So wouldn’t
it be more sensible to regard these two groups as algebraically the same,
the only differences being purely cosmetic? Indeed, doesn’t any cyclic
group of order 6 look like
{
e, x, x
2
, x
3
, x
4
, x
5
, x
6
}
?
236
CHAPTER 4
Abstract Algebra
Here’s a much less obvious example. Consider the two infinite groups
(
R
,
+) and (
R
+
,
·
).
At first blush these would seem quite different.
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