# Prove a subgroup of G/H X G/K is isomorphic to G/(H intersect K)

1. Nov 6, 2012

### Avatarjoe

1. The problem statement, all variables and given/known data

Suppose H and K are normal subgroups of G. Prove that G/H x G/K has a subgroup isomorphic to G/(H$\cap$K)

2. Relevant equations

3. The attempt at a solution

I was trying to find a homomorphism from G to G/H x G/K where G/(H$\cap$K) is the kernal. Maybe something like if g is in H it getts mapped to (Hg, e), but nothing like that worked. I'm really stuck on this one.

2. Nov 6, 2012

### jbunniii

That doesn't make any sense, because $G/(H \cap K)$ is not a subgroup of $G$. What you need is a homomorphism with kernel $H \cap K$.

Assuming you're working with right cosets, the natural homomorphism from G to G/H is $g \mapsto Hg$, and the natural homomorphism from G to G/K is $g \mapsto Kg$. Can you use these to construct a homomorphism from G to G/H x G/K?

3. Nov 7, 2012

### Avatarjoe

Right. Thanks, that is what I meant. So g↦ (Hg, Kg) is a homomorphism and that's easy to prove. Also H∩K will be the kernal. If I'm not mistaken, I just need to show that {Hg x Kg} is a subgroup of G/H x G/K and then the homomorphism will be onto that subgroup by the way I defined it. Then G/(H∩K) is isomorphic to {Hg x Kg}

4. Nov 7, 2012

### jbunniii

No need to do that - surely you have encountered the theorem that the image of a homomorphism is always a subgroup. If not, it's easy to prove - easier than trying to show it for a specific case like this one.

By the way, I understand what you mean by your notation, {Hg x Kg}, but I don't think it's very good notation. I would suggest writing something like $\{(Hg, Kg) : g \in G\}$.

5. Nov 7, 2012

### micromass

Another question: have you seen the isomorphism theorems yet?? They might come in handy.

6. Nov 8, 2012

### jbunniii

I have been implicitly assuming that he has the first isomorphism theorem available. Avatarjoe, is that a valid assumption?