Page 61 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes So let us start with a definition.
Definition: The order of a finite group G is the number of elements in G. It is denoted by o(G).

For example, o(S ) = 6 and o(A ) = 3. Remember, A = {I, (1 2 3), (1 3 2)}!
3
3
3
You can also see that o(Z ) = n. And, you know that o(S ) = n!.
n
n
Now, let G be a finite group and H be a subgroup of G. We define a function f between the set of
right cosets of H in G and the set of left cosets of H in G by
f : { Hx | x  G }  { yH | y  G } : f(Hx) = x H.
–1
Definition: Let H be a subgroup of a finite group G. We call the number of distinct cosets of H in
G the index of H in G, and denote it by | G : H |.
Thus, we see that | S : A | = 2.
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Note that, if we take H = {e}, then | G : {e} | = o(G), since {e}g = {g)  g  G and {e)g  {e}g’
if g  g’.
Now let us look at the order of subgroups. In last unit you saw that the orders of the subgroups
of S are 1, 2, 3 and 6. All these divide o(S ) = 6. This fact is part of a fundamental theorem about
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finite groups. Its beginnings appeared in a paper in 1770, written by the famous French
mathematician Lagrange. He proved the result for permutation groups only. The general result
was probably proved by the famous mathematician Evariste Galois in 1830.
Theorem 3 (Lagrange): Let H be a subgroup of a finite group G. Then

o(G) = o(H) | G : H |. Thus, o(H) divides o(G) and | G : H | divides o(G).
Proof: You know that we can write G as a union of disjoint right cosets of H in G. So, if Hx , Hx ,
2
l
...., H , are all the distinct right cosets of H in G, we have
x
G = Hx  Hx  ...  Hx ( 1 )
1 2 r
and | G : H | = r.
We know that | Hx | = | Hx | = ... = | Hx | = o(H).
r
1
2
Thus, the total number of elements in the union on the right hand side of ( I ) is
o(H) + o(H) + .....+ o(H) (r times) = r o(H).
Therefore, (1) says that o(G) = r o(H)
= o(H) | G : H |.

You will see the power of Lagrange’s theorem when we get down to obtaining all the subgroups
of a finite group.
For example, suppose we are asked to find all the subgroups of a group G of order 35. Then the
only possible subgroups are those of order 1, 5, 7 and 35. So, for example, we don’t need to waste
time looking for subgroups of order 2 or 4.

In fact, we can prove quite a few nice results by using Lagrange’s theorem. Let us prove some
results about the order of an element.
Definition: Let G be a group and g  G. Then the order of g is the order of the cyclic subgroup
< g >, if < g > is finite. We denote this finite number by o(g). If < g > is an infinite subgroup of G,
we say that g is of infinite order.

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