# Let $M$ and $N$ be two normal subgroups of $G$. Show that $M \cap N$ is also normal in $G$. Then show that $G/(M \cap N) \cong (G/M) \times (G/N)$

Let $M$ and $N$ be two normal subgroups of $G$. Show that $M \cap N$ is also normal in $G$. Furthermore,if $G=MN$ then show that $$G/(M \cap N) \cong (G/M) \times (G/N)$$ I manage to prove the first part. But for the two quotient groups to be isomorphic, I have no idea. Anyone can help ?

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The second assertion is true if we add one more condition that $G=MN$, if we define the map $$f:G\to (G/M) \times (G/N)$$ such that $$f(g)=(gM,gN)$$ then $f$ is a homomorphism and it is easily seen that $kerf=M\cap N$, now if we can prove that $f$ is onto we can use the first isomorphism theorem to get the result.

will let $(xM,yN)\in (G/M) \times (G/N)$ now since $G=MN$ we have $x=mn$ and $y=m'n'$ but then we have$$xM=mnM=n(n^{-1}mn)M=nM$$ since $m\in M$ and $M$ is normal,similarly we get$$yN=m'N$$ now consider $$f(m'n)=(m'nM,m'nN)=(n(n^{-1}m'n)M,m'nN)=(nM,m'N)=(xM,yN)$$ again since $m\in M$ and $M$ is normal and hence that map is onto and we have $$G/(M \cap N) \cong (G/M) \times (G/N).$$

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More generally we have $G/(M \cap N) \cong G/M \times_{G/(NM)} G/N$ (fiber product). – Martin Brandenburg Feb 21 '13 at 19:41
Is the multivariable statement also true: if $N_1,\ldots,N_n\unlhd G$ and $N_iN_j=G$ for all $i<j$, then $$\frac{G}{N_1\cap\ldots\cap N_n}\cong\frac{G}{N_1}\times\ldots\times\frac{G}{N_n}?$$ @MartinBrandenburg – Leon Feb 7 at 21:52
@Leon: No. (Consider vector spaces, count dimension.) – Martin Brandenburg Feb 7 at 22:11
Wow, you are right, thanks!!! mathoverflow.net/questions/17740/… – Leon Feb 8 at 20:00

Second assertion is not true. E.g., for $M=N=1$ you get from your isomorphism $G\cong G\times G$.

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There is nothing inherently wrong with $G\times G\cong G$. (It is impossible for finite groups though, of course. ) – rschwieb Feb 21 '13 at 17:33
+1. Just say that you assume $2\leq |G|<+\infty$. – 1015 Feb 21 '13 at 17:35

Part two is false: $6\Bbb Z\cap 3\Bbb Z=6 \Bbb Z$, but $\Bbb Z /6\Bbb Z$ is not isomorphic to $\Bbb Z /3\Bbb Z\times \Bbb Z /6\Bbb Z$

The former has 6 elements, but the latter has 18.

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Can you think of a homomorphism from the left side to the right side? In particular if $[g]$ represents the equivalence class of some $g \in G$ on the left side, where should it be sent to on the right side? Can you prove that this is well-defined? An injection? A surjection?

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Probably the task is to show that $M \cap N$ is normal and that $G/(M \cap N)$ embeds into $G/M \times G/N$ (do you really have reproduced the exercise correctly?). In fact, both can be done in one step: Consider the diagonal map $G \to G/M \times G/N$, $g \mapsto ([g],[g])$. It is a homomorphism, with kernel $M \cap N$. Done.

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what is diagonal map ? do we need to show that it is well-defined ? – Idonknow Feb 21 '13 at 18:01
the diagonal map is the map Martin defined above, that sends $g\to (g+M,g+N)$, the kernel of this map is $M\cap N$ but this map is not onto so you can't get $G/(M \cap N) \cong (G/M) \times (G/N)$ from this map, but you can get $G/(M \cap N)$ is isomorphic to a subgroup of $(G/M) \times (G/N)$. – i.a.m Feb 21 '13 at 18:43