# Thread: How do you show that this is abelian?

1. ## How do you show that this is abelian?

Show that if H,K are normal to G and H intersect K = {1}, then hk=kh for all h in H, and k in K.

I have tried playing around with this for a long time. I just can't quite get it right. I've used the facts of the normal subgroup tests, gHg^-1 in H and gKg^-1 in K, the fact that gH=Hg and gK=Kg. I've tried using kHk^-1 in H and hKh^-1 in K. I just can't quite get it to work. All I am able to achieve is getting hk=k2h and kh=h2k...

Any tips would be greatly appreciated!!! I have tried searching around a bit and can't find anything.

2. Originally Posted by xoyankeegirlx3
Show that if H,K are normal to G and H intersect K = {1}, then hk=kh for all h in H, and k in K.

I have tried playing around with this for a long time. I just can't quite get it right. I've used the facts of the normal subgroup tests, gHg^-1 in H and gKg^-1 in K, the fact that gH=Hg and gK=Kg. I've tried using kHk^-1 in H and hKh^-1 in K. I just can't quite get it to work. All I am able to achieve is getting hk=k2h and kh=h2k...

Any tips would be greatly appreciated!!! I have tried searching around a bit and can't find anything.
$h^{-1}k^{-1}hk=h^{-1}\left(k^{-1}hk\right)\in H$ , since H is normal, but also $h^{-1}k^{-1}hk=\left(h^{-1}k^{-1}h\right)k\in K$ ,

because K is normal, thus...

Tonio

3. Originally Posted by xoyankeegirlx3
Show that if H,K are normal to G and H intersect K = {1}, then hk=kh for all h in H, and k in K.

I have tried playing around with this for a long time. I just can't quite get it right. I've used the facts of the normal subgroup tests, gHg^-1 in H and gKg^-1 in K, the fact that gH=Hg and gK=Kg. I've tried using kHk^-1 in H and hKh^-1 in K. I just can't quite get it to work. All I am able to achieve is getting hk=k2h and kh=h2k...

Any tips would be greatly appreciated!!! I have tried searching around a bit and can't find anything.
Also, one can show that $\phi:H\times K\to Gh,k)\mapsto hk" alt="\phi:H\times K\to Gh,k)\mapsto hk" /> is an isomorphism--this tells you more than you need though.

4. Originally Posted by Drexel28
Also, one can show that $\phi:H\times K\to Gh,k)\mapsto hk" alt="\phi:H\times K\to Gh,k)\mapsto hk" /> is an isomorphism--this tells you more than you need though.

The above is an isomorphism iff $G=HK$ , otherwise it is only a monomorphism.

Tonio

5. Originally Posted by tonio
The above is an isomorphism iff $G=HK$ , otherwise it is only a monomorphism.

Tonio
Tonio,

I did not see that this condition was not included. My mistake.

6. Originally Posted by Drexel28
Tonio,

I did not see that this condition was not included. My mistake.

What?!!? You oversaw, or added on your own, a condition? Welcome to the club...

Tonio