How can I prove that $G/M\cap N \cong M/(M\cap N) \times N/(M\cap N)$ where $M,N \triangleleft G$ and $M.N=G$?

Question

If $M\cap N = \{e\}$ then $M/M\cap N = M$, $N/M\cap N = M$ and $G/M\cap N = G$ and $G = M\times N$ trivially by the def of direct product.

If $M\cap N \neq \{e\}$, I was trying to follow the same line of argument , to show that $M\cap N \triangleleft G$ and hence $M\cap N \triangleleft M,N$.

Now, I think $a(M\cap N) \neq b(M\cap N)$ $ \forall a \in M\setminus M\cap N $ and $ \forall b \in N\setminus M\cap N $ and therefore
$M/(M\cap N) \cap N/(M\cap N) = \{e\}$

but I am not able to show $(M/M\cap N) \triangleleft (G/M\cap N)$ because this happens
$\iff a(M/M\cap N) = (M/M\cap N)a$ $ \forall a\in G/M\cap N$
$\iff g(M\cap N)m(M\cap N) = m(M\cap N)g(M\cap N)$
$\iff gm(M\cap N) = mg(M\cap N)$
$\iff gm = mg$
which obviously need not be true. Can you please suggest a modification to my approach or a new approach altogether? Appreciate your response.

Answer 1

Let me write $K=M\cap N$. In your list of equivalences, you make the step $$gmK = mgK \qquad\Leftrightarrow\qquad gm=mg.$$ Though it is unclear what $g$ and $m$ are, this is not true in general. The implication '$\Leftarrow$' is of course true, but the implication '$\Rightarrow$' is false; it does imply that $gmg^{-1}m^{-1}\in K$, for example.

Because $M,N\triangleleft G$, for all $g\in G$ we have $gMg^{-1}=M$ and $gNg^{-1}=N$. Of course $K\subset M$ and $K\subset N$, so

$$gKg^{-1}\subset M\qquad\text{ and }\qquad gKg^{-1}\subset N,$$ which implies that $gKg^{-1}\subset M\cap N=K$. Because multiplication by $g$ and $g^{-1}$ is bijective, it follows that $gKg^{-1}=K$, so $K\triangleleft G$.

Showing that $M/K\triangleleft G/K$ is now quickly done; now that you know $K\triangleleft G$ the natural map $$G/K\ \longrightarrow\ G/M,$$ is well-defined. It is surjective because $K\subset M$, so you can use the first isomorphism theorem.

Answer 2

Hint/oft encountered approach: Recall that for a group homomorphism $\varphi : G \to H$ we have

$$G/\ker \varphi \cong \operatorname{Im} \varphi.$$

Pick $G = M \times N$ and $H = M/(M\cap N) \times N/(M\cap N)$ and $\varphi : G \to H$ the "obvious" map (I'll leave it to you to decide whether or not it is obvious).