Normal Subgroup

1. Nov 16, 2007

sam09

Hi, I need some help with this problem:

Let H be a subgroup of an arbitary group G. Prove H is normal iff it has the following property: For all a,b, in G, ab is in H iff ab is in H.

2. Nov 16, 2007

d_leet

Are you sure this is the exact question? Did you notice that this is actually true for all subgroups of G, and is equivalent to saying c is in H if and only if c is in H?

3. Nov 16, 2007

sam09

Re

oops, I meant to write for all a,b in G, ab is in H iff ba is in H.

4. Nov 16, 2007

matt grime

It is a prerequisite of the forums that you say what you're trying to do to solve the problem. Start with the definition of normal.

5. Nov 16, 2007

Jim Kata

I'll do half of the proof the other is the same, but with the letters reversed.

if ab is an element of H and H is a normal group then

abab = h for some h element of H

ba = (a^-1)h(b^-1)

gbag^-1 = g(a^-1)h(b^-1)g^-1 for any element g of G

gbag^-1 = g(a^-1)(abb^-1a^-1)h(b^-1)g^-1

but since ab is an element of H so is b^-1a^-1 since H is a group

so
gbag^-1 = (gb)h'(gb)^-1 for h' = b^-1a^-1h which is also an element of H

now by definition of a normal group (gb)h'(gb)^-1 = h'' for some element of H

Therefore gbag^-1 = h'' some element of H

That completes one direction the other direction can be done by the same method.