Normal Subgroups intersection = <e>

[kathrynag]

Let H and K be normal subgroups of G such that H intersect K=<e>. Show that hk=kh for all h in H and k in K.
H and K are normal so ghg^-1 is in H and gkg^-1 is in K.
want to show hk=kh. So basically I'm showing this is abelian.
Can I do ghg^-1=gkg^-1?
ghg^-1g=gkg^-1g
gh=gk
so that works if g=h

 

[wnorman27]

Let $h\in\ H$ with $h\neq e$, and let $k\in\ K$ with $k\neq e$.

Then what can you say about $khk^{-1}$?

 

[I like Serena]

Let H and K be normal subgroups of G such that H intersect K=<e>. Show that hk=kh for all h in H and k in K.
H and K are normal so ghg^-1 is in H and gkg^-1 is in K.
want to show hk=kh. So basically I'm showing this is abelian.
Can I do ghg^-1=gkg^-1?
ghg^-1g=gkg^-1g
gh=gk
so that works if g=h

Hi kathrynag!

I don't see how what you did could be correct.

When showing commutativity it is often useful to look at the commutator.
The commutator of h and k is [h,k]=hkh^-1k^-1.
If the commutator of h and k is equal to e, than h and k commute (why?).

Suppose you write hk=k₁h, which should be true for some k₁ in K.
What do you get?

 

[I like Serena]

Oh, I just noticed that this is a pretty old thread.
I suspect that the OP is not interested in the answer anymore.