Proving H is Normal: Let G be an Arbitrary Group

[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.

 

[d_leet]

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.

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?

 

[sam09]

Re

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

 

[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.

 

[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 Ggbag^-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.