Is H Always a Normal Subgroup in a Group of Even Order?

[cmurphy]

Let G be a finite group of even order with n elements. H is a subgroup of G, with n/2 elements.

I need to show that H is normal. I have set up the phi function to be: phi(x) = 1 if x is an element of H. phi(x) = -1 if x is not an element of H. Thus H is the kernal.

I am trying to show that H is a homomorphism. Then from that I know that H is normal.

I am breaking this up into cases. I have been successful showing that the homomorphism holds if x,y are both in H. It also holds if x is in H and y is in G.

However, I am having difficulty showing that if x and y are both in G, then their product xy must be in H. How do I go about showing that?

Colleen

 

[matt grime]

The thing you're trying to show is false, so I'd stop trying to show it.

Just consider the cosets G/H and let G act on them. Let the cosets be [e] and [x], show that the action of G on them by multiplication gives a homomorphism to C_2 and its kernel is H. Remember cosets are equal or disjoint.

 

[M98Ranger]

,

To show that H is normal, we need to prove that for any element x in G and any element h in H, the conjugate of h by x (i.e. xhx^-1) is also in H. This means that H is closed under conjugation by elements of G, which is a key property of normal subgroups.

In order to show this, we can use the fact that H has n/2 elements and G has n elements. Since G has even order, we can write n = 2m for some positive integer m. This means that H has m elements, and G has 2m elements.

Now, consider the product xy, where x and y are both in G. Since G has 2m elements, there are two cases: either x and y are both in H, or one of them is in H and the other is not.

In the first case, where x and y are both in H, we know that their product xy is also in H because H is a subgroup and is closed under multiplication.

In the second case, where one of x or y is in H and the other is not, let's say without loss of generality that x is in H and y is not. This means that phi(x) = 1 and phi(y) = -1. Now, consider the product phi(x)phi(y). Since phi(x) = 1 and phi(y) = -1, we have phi(x)phi(y) = 1(-1) = -1. But, since x and y are both in G, their product xy must also be in G. This means that phi(xy) = -1. But, since H has m elements and G has 2m elements, we know that phi(xy) = 1, because there are exactly m elements in H and 2m elements in G that are not in H (since G has n elements and H has n/2 elements). This is a contradiction, since we have shown that phi(xy) = -1 and phi(xy) = 1. Therefore, our assumption that one of x or y is in H and the other is not must be false, and we can conclude that xy is in H.

Hence, we have shown that for any elements x in G and h in H, the conjugate of h by x (i.e. xhx^-1) is