# Normal and Simple groups

1. Nov 28, 2009

### 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.

2. Nov 28, 2009

### morphism

What if G=H?

3. Nov 29, 2009

### Bleys

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

4. Nov 29, 2009

### morphism

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

5. Nov 29, 2009

### 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...

6. Nov 29, 2009

### 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.

7. Nov 29, 2009

### 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?