# Two normal subgroups and isomorphism theorem [duplicate]

Question

Let $N_1$ and $N_2$ be normal subgroups of $G$.

Prove that $N_1N_2/(N_1\cap N_2) \cong (N_1N_2/N_1)\oplus (N_1N_2/N_2)$.

I think the homomorphism must be $\phi : N_1N_2 \to (N_1N_2/N_1)\oplus (N_1N_2/N_2)$

such that $\phi(n_1n_2)=(n_1n_2N_1,n_1n_2N_2)$

Then by 1st isomorphism theorem, $N_1N_2/ker\phi \cong Im\phi$

and $ker\phi = N_1 \cap N_2$

But it is hard to show that $\phi$ is surjective.

What should I do?

## marked as duplicate by Arnaud D., YiFan, Alexander Gruber♦ abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 29 at 23:04

• Your subgroups are normal, so your homomorphism is actually $\phi(n_1 n_2) = (n_2 N_1, n_1 N_2)$. The surjectivity should be obvious now. – Starfall May 6 '16 at 15:58
• You need to be careful and show that the $n_1n_2$ representation is well-defined. – Steve D May 6 '16 at 16:25
• Here's another way to think about it: use the second isomorphism theorem on the two factors on the right. – Steve D May 6 '16 at 16:25
• @SteveD Thanks for noting well-defined – Pearl May 6 '16 at 23:07

As you note, it is a simple application of the first isomorphism theorem.

You have correctly defined a morphism $$\varphi : N_{1} N_{2} \to \dfrac{N_1 N_2}{N_1} \times \dfrac{N_1 N_2}{N_2}$$ by $$\varphi(g) = (g N_{1}, g N_{2}),$$ and checked that the kernel is $N_{1} \cap N_{2}$.

As to the image, take an arbitrary element of $$\dfrac{N_1 N_2}{N_1} \times \dfrac{N_1 N_2}{N_2}$$ and rewrite it $$(n_{1} n_{2} N_{1}, n_{1}' n_{2}' N_{2}) = (n_{2} N_{1}, n_{1}' N_{2}) = (n_{1}' n_{2} N_{1}, n_{1}' n_{2} N_{2}) = \varphi(n_{1}' n_{2}),$$ where $n_{i}, n_{i}' \in N_{i}$.

You can assume $N_1\cap N_2=\{1\}$, by working in $G/(N_1\cap N_2)$ and using the fact that $$N_1N_2/N_1\cong(N_1N_2/(N_1\cap N_2))\big/(N_1/(N_1\cap N_2))$$ and similary for the quotient modulo $N_2$. It is also not restrictive to assume $G=N_1N_2$.

In this case the isomorphism becomes the usual proof that, for a group $G$ with two normal subgroup $A$ and $B$ such that $AB=G$ and $A\cap B=\{1\}$, we have $G\cong A\times B$.

Here's a quick proof:

The second isomorphism theorem shows

$$N_1N_2/(N_1)\cong N_2/(N_1\cap N_2)$$

and similarly for the other right-hand group.

So it would be enough to define a map $N_1\oplus N_2\rightarrow N_1N_2/(N_1\cap N_2)$ whose kernel is $N_1\cap N_2$. That should be easy.

• I use the homomorphism $N_1 \oplus N_2 \to N_1N_2/(N_1\cap N_2)$ by $(n_1,n_2)$ to $n_1n_2(N_1 \cap N_2)$. But it is hard to show that kernel is $N_1 \cap N_2$ – Pearl May 6 '16 at 22:40
• @pearl: might be easier to consider each map in turn: so first look at the map $N_1\rightarrow N_1N_2/(N_1\cap N_2)$, and think about the kernel there. Do the same then for the $N_2$ side. – Steve D May 7 '16 at 0:31