Page 105 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes Similarly, (1 3 2) = (1 2 3)  H. By the same reasoning (1 4 2), (1 2 4), (1 4 3), (1 3 4), (2 3 4), (2 4 3)
2
are also distinct element of H. Of course, I  H.
Thus, H contains at least 9 elements.
 o(H)  9. This contradicts our assumption that o(H) = 6.
Therefore, A has no subgroup of order 6.
4
We use A to provide another example too. (See how useful A is!) In earlier unit we’d said that
4
4
if H  N and N  G, then H need not be normal in 6. Well, here’s the example.’
Consider the subset V = {I, (1 2) (3 4), (1 4) (2 3), (1 3) (2 4)) of A .
4
4
Now, let H = {I, (1 2) (3 4)}. Then H is a subgroup of index 2 in V .  H A V .
4
4
So, H  V , V  A . But H  A . Why? Well, (1 2 3)  A is such that
4
4
4
4
4
(1 2 3) (1 2) (3 4) (1 2 3) = (1 3) (2 4)  H.
-1
And now let us see why permutation groups are so important in group theory.
8.4 Cayley’s Theorem
Most finite groups that first appeared in mathematics were groups of permutations. It was the
English mathematician Clayley who first realised that every group has the algebraic structure
of a subgroup of S(X), for some set X. In this section we will discuss Cayley’s result and some of
its applications.
Theorem 7 (Cayley): Any group G is isomorphic to a subgroup of the symmetric group S(G).

Proof: For a  G, we define the left multiplication function
f : G  G : f (x) = ax.
4
a
f is 1-1, since
a
fa(x) = fa(y)  ax = ay  x = y x, y E G.
fa is onto, since any x E G is f, (a – ’x).
:. f  S(G)  a  G.
a
(Note that S(G) is the symmetric group on the set G.)

Now we define a function f : G  S(G) : f(a) = fa.
We will show that f is an injective homomorphism. For this we note that

(f ,f ) (x) = f (bx) = abx = f (x)  a, b  G.
ao
a
ab
b
 f(ab) = f = f o f = f(a) of (b)  a, b  G.
ab
a
b
That is, f is a homomorphism.
Now, Ker f = (a  G | fa = I )
G
= ( a  G | f (x)=x  x  G }
a
= ( a  G |a x = x  x  G }
= {e}.

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