# Is this a surjective homomorphism?

1. Oct 9, 2008

### alligatorman

I'm trying to prove that if M,N are normal in G and MN = G, then

$$G/(M\cap N)\cong G/M \times G/N$$

In an attempt to use the 1st Isom. Thm, I have a homomorphism from G to G/M x G/N :

$$g \mapsto (gM, gN)$$

The kernel is $$M\cap N$$, so I just have to show that the function is onto to get the isomorphism.

My guess is that (aM, bN) = (abM, abN). I am having a difficult time showing this, or I may be wrong.

Any help?

Last edited: Oct 9, 2008
2. Oct 9, 2008

### morphism

The fact that G=MN=NM is important here. Write a=n1m2 and b=m2n2 and expand.

3. Oct 9, 2008

### alligatorman

I don't understand how if you have a=n_1 m_2, then you can write b=m_2 n_2, if a and b are arbitrary

4. Oct 9, 2008

### morphism

Sorry, that was an unfortunate typo. It should've been "a=n1m1" instead.

5. Oct 9, 2008

### alligatorman

So suppose $$a=n_1m_1$$ and $$b=m_2n_2$$

Then $$aM=n_1 M$$ and $$bN=m_2 N$$.

Also, $$abM=n_1m_1m_2n_2M=n_1m_3n_2M=n_1n_2m_4M=n_3M$$

I've been trying this method for days but I can't connect the two. I must be missing some important fact about cosets and/or normal groups

6. Oct 9, 2008

### morphism

Here's another hint: $(n_1 M, m_2 N) = (n_1 M, n_1 N) (m_2 M, m_2 N)$.