Normal and Simple Subgroups in Finite Group G: Proof of Equality for K and H

[Bleys]

H,K are normal subgroups of a (finite) group G, and K is also normal in H. If G/K and G/H are simple, does it follow that H=K?
I'm almost convinced it does, but I'm having trouble proving it. I mean, the cosets of H partition G and the cosets of K partition G in the same way and on top of that partition H, right? I'm not sure when to bring in normality and the fact that the quotients are simple.

 

[morphism]

What if G=H?

 

[Bleys]

no, sorry, I should have included that. H and K are proper normal subgroups in G.

 

[morphism]

Okay. Now, can you describe the relationship(s) between G/H, G/K and H/K?

 

[Bleys]

This is actually where I had trouble.
$$G/H = \left\{ H, g_{1}H, g_{2}H,... \right\}$$ and $$G/K = \left\{ K, g_{1}K, ... \right\} = \left\{ K, g_{1}K, ... \right\} \cup \left\{ K, h_{1}K, ... \right\}$$ for $$g_{i} \in G-H$$ and $$h_{j} \in H$$
That last on is $$G/K = \left\{ K, g_{1}K, ... \right\} \cup H/K$$ for $$g_{i} \in G-H$$. Then, can you say that $$G/H = \left\{ K, g_{1}K, ... \right\}$$ for $$g_{i} \in G-H$$, since the intersection of H/K and $$\left\{ K, g_{1}K, ... \right\}$$ is just the identity element, K. Then the cosets of H in G must coincide with the cosets of K in G. Thus H=K?
Or am I totally missing it. I must be, since I haven't used either normality and simple quotients...

 

[morphism]

I have to run, so I'll leave you with a quick exercise (or reminder, in case you've seen this result previously): (G/K)/(H/K) = G/H. This isomorphism pretty much solves your problem. One way to prove it is to write down an obvious map G/K -> G/H and look at its kernel.

 

[Bleys]

oh alright. So the proof of the isomorphism uses normality of K in G and H in G (specifically for the map to be a group homomorphism) and the First Isomorphism Theorem. Now, the kernel of the map is normal in its domain (ie G/K), so ker = H/K is normal in G/K. But G/K is simple, so either H/K = {1} or H/K = G/K. If H/K = G/K then (G/K)/(H/K) = {1} and so G/H = {1}. But this implies G=H, which is not the case. Therefore H/K = {1}, and so H=K. Great, thanks a lot!

But one more thing. I didn't use the fact the G/H is simple. Does it mean you can relax that condition to just exclude it?