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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 Apr 29 at 23:04

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  • $\begingroup$ 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. $\endgroup$ – Starfall May 6 '16 at 15:58
  • 1
    $\begingroup$ You need to be careful and show that the $n_1n_2$ representation is well-defined. $\endgroup$ – Steve D May 6 '16 at 16:25
  • $\begingroup$ Here's another way to think about it: use the second isomorphism theorem on the two factors on the right. $\endgroup$ – Steve D May 6 '16 at 16:25
  • $\begingroup$ @SteveD Thanks for noting well-defined $\endgroup$ – 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.

  • $\begingroup$ 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 $N_1 \cap N_2$ $\endgroup$ – Pearl May 6 '16 at 22:40
  • $\begingroup$ @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. $\endgroup$ – Steve D May 7 '16 at 0:31

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