Is there an isomorphism from G to (G/M)x(G/N) with the kernel M\bigcapN?

[symbol0]

Let M and N be normal subgroups of G such that G=MN.
Prove that G/(M$\bigcap$N)$\cong$(G/M)x(G/N).

I tried coming up with an isomorphism from G to (G/M)x(G/N) such that the kernel is M$\bigcap$N, so that I can use the fundamental homomorphism theorem.
I tried f(a) = (aM, aN). It is an homomorphism and M$\bigcap$N is the kernel but I'm having a hard time showing it is onto.

I would appreciate any help.
Thank you

 

[CompuChip]

So you need to show that for all g, g' in G, there is an a in G such that: (g M, g' N) = (a M, a N).

I haven't worked this out in detail, but: since G = MN, you can write g = m n, g' = m' n'. I suspect that a = m' n might do the trick.
You will need that M and N are normal, so in particular h M = M h, h N = N h for all h in G.

 

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Thank you Compuchip