Page 70 - DMTH403_ABSTRACT_ALGEBRA

Unit 5: Normal Subgroups

In above Example, both < a > and < b > are of index 2 in K . We have the following result about Notes
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such subgroups.
Theorem 3: Every subgroup of a group G of index 2 is normal in G.
Proof: Let N  G such that | G : N | = 2. Let the two right cosets of N be N and Nx, and the two
left cosets be N and yN.
Now, G = N  yN, and x  G.  x  N or x  yN.

Since N  Nx = , x  N.  x  yN, xN = yN.
To show that N  G, we need to show that Nx = xN.

Now, for any n  N, nx  G = N  xN. Therefore, nx  N or nx  xN.
But nx  N, since x  N.  nx  xN.
Thus, Nx  xN.

By a similar argument we can show that xN  Nx.
 Nx = xN, and N  G.

We will use this theorem to show that, for any n  2, the alternating group A, is a normal
subgroup of S .
n
In fact, if you go back b, you can see that A  S , since Lagrange’s theorem implies that
4
4
o(S ) 4!
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|S : A | o(A )  12  2.
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4
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Consider the quaternion group Q , which we discussed earlier. It has the following 6 subgroups:
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Ho = (I}, H = {I, – I}, H = (I, – I, A, – A), H = {I, – I, B, – B},
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3
I
H = {I, – I, C, – C), H = Q .
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8
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You know that H and H are normal in Q . Using Theorem 3, you can see that Hz, H and H are
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0
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5
3
normal in Q . e
By actual multiplication you can see that
g
1
g H g  H   Q .  H  Q .

1
1
8
1
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Therefore, all the subgroups of Q are normal.
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But, you know that Q is non-abelian (for instance, AB = – BA).
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So far we have given examples of normal subgroups. Let us look at an example of a subgroup
that isn’t normal.
Example: Show that the subgroup < (1 2) > of S is not normal.
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Solution: We have to find g  S such that g (1 2)g  < (1 2) >.
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Let us try g = (1 2 3).
Then, g (l 2)g = (3 2 1) (1 2) (1 2 3)
-1
= (3 2 1) (2 3) = (1 3)  < (1 2)>
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