Exercise 3.3.7 (If $G = MN$ ($M \unlhd G$, $N \unlhd G$) then $G/(M\cap N) \simeq (G/M) \times (G/N)$)

Proof. Since $N⊴G$, by the Second Isomorphism Theorem, $M\cap N⊴M$, and

Similarly, since $M⊴G$,

Now we show that $\bar{G}=G∕M\cap N$ is isomorphic to the direct product of $\bar{M}=M∕M\cap N$ and $\bar{N}=N∕M\cap N$.

(We write $\bar{H}=H∕M\cap N$ if $M\cap N\le H$, and $\bar{g}=g(M\cap N)$ the coset of $g\in G$ relative to $M\cap N$.)

The Lattice Isomorphism Theorem shows that

that is

Moreover, since $G=\mathit{𝑀𝑁}$, every $\bar{g}\in \bar{G}$ is a coset $\bar{g}=g(M\cap N)$, where $g\in G=\mathit{𝑀𝑁}$, so $g=\mathit{𝑚𝑛},m\in M,n\in N$, and since $M\cap N⊴G$,

This shows that $\bar{G}\subseteq \bar{M}\bar{N}$, where $\bar{M}\le \bar{G}$, $\bar{N}\le \bar{G}$, so

Moreover, since $M⊴G$ and $N⊴G$, the part (d) of the Lattice Isomorphism Theorem shows that $\bar{M}⊴\bar{G}$ and $\bar{N}⊴\bar{G}$.

We know that

(i)

$\bar{M}⊴\bar{G}$, $\bar{N}⊴\bar{G}$,

(ii)

$\bar{G}=\bar{M}\bar{N}$,

(iii)

$\bar{M}\cap \bar{N}=\{\bar{1}\}$.

This is sufficient to show that $\bar{G}\simeq \bar{M}\times \bar{N}$ (this is a classic caracterisation of the direct product):

We show first that every element $\bar{m}\in \bar{M}$ commutes with every element $\bar{n}\in \bar{N}$. Using (i),

Therefore $\bar{m}\bar{n}{\bar{m}}^{-1}{\bar{n}}^{-1}\in \bar{M}\cap \bar{N}=\{\bar{1}\}$ (by (iii)), so

Consider now the map

Then

• $\phi$ is a homomorphism: If $(\bar{m},\bar{n})\in \bar{M}\times \bar{N}$ and $(\overline{m^{\prime }},\overline{n^{\prime }})\in \bar{M}\times \bar{N}$, then ,

  $$\begin{array}{llllll}\hfill \phi (\bar{m},\bar{n})\phi (\overline{m^{\prime }},\overline{n^{\prime }})& =\bar{m}\bar{n}\overline{m^{\prime }}\overline{n^{\prime }}& \hfill & & \hfill & \\ \hfill & =\bar{m}\overline{m^{\prime }}\bar{n}\overline{n^{\prime }}& \hfill (\text{since }\bar{n}\overline{m^{\prime }}=\overline{m^{\prime }}\bar{n})& & \hfill & & \hfill \\ \hfill & =\phi (\bar{m}\overline{m^{\prime }},\bar{n}\overline{n^{\prime }})& \hfill & & \hfill & \\ \hfill & =\phi ((\bar{m},\bar{n})(\overline{m^{\prime }},\overline{n^{\prime }})).& \hfill & & \hfill & \end{array}$$
• $\phi$ is surjective: By (ii), $\bar{G}=\bar{M}\bar{N}$, so $\phi$ is surjective.

• $\phi$ is injective: For all $(\bar{m},\bar{n})\in \bar{M}$,

  $$(\bar{m},\bar{n})\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )⟺\bar{m}\bar{n}=\bar{1}⟺\bar{m}={\bar{n}}^{-1},$$

  and since $\bar{m}={\bar{n}}^{-1}\in \bar{M}\cap \bar{N}=\{\bar{1}\}$, we obtain $(\bar{m},\bar{n})=(\bar{1},\bar{1})$, so $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=\{\bar{1},\bar{1}\}$ is trivial. This shows that $\phi$ is injective.

Therefore $\phi$ is an isomorphism, so

that is

Then using the isomorphisms (1) and (2), we obtain

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