# Internal Group Direct Product Isomorphism

## Theorem

Let $G$ be a group.

Let $H_1, H_2$ be subgroups of $G$.

Let $\phi: H_1 \times H_2 \to G$ be the mapping defined by $\map \phi {h_1, h_2} := h_1 h_2$.

If $\phi$ is a (group) isomorphism, then both $H_1$ and $H_2$ are normal subgroups of $G$.

## Proof

$\phi$ is an isomorphism, so in particular a (group) homomorphism.

Thus by Induced Group Product is Homomorphism iff Commutative, every element of $H_1$ commutes with every element of $H_2$.

Now suppose $a \in G$.

As $\phi$ is an isomorphism, it follows that $\phi$ is surjective.

$\exists h_1 \in H_1, h_2 \in H_2: a = h_1 h_2$

Now any element of $a H_1 a^{-1}$ is in the form $a h a^{-1}$ for some $h \in H_1$.

Thus:

 $\displaystyle a h a^{-1}$ $=$ $\displaystyle \paren {h_1 h_2} h \paren {h_1 h_2}^{-1}$ $\displaystyle$ $=$ $\displaystyle h_1 h_2 h h_2^{-1} h_1^{-1}$ $\displaystyle$ $=$ $\displaystyle h_1 h h_2 h_2^{-1} h_1^{-1}$ $h \in H_1$ and $h_2 \in H_2$ commute $\displaystyle$ $=$ $\displaystyle h_1 h h_1^{-1} \in H_1$

Thus $a H_1 a^{-1} \subseteq H_1$, and $H_1$ is therefore a normal subgroup of $G$.

Similarly, any element of $a H_2 a^{-1}$ is in the form $a h a^{-1}$ for some $h \in H_2$.

Thus:

 $\displaystyle a h a^{-1}$ $=$ $\displaystyle \paren {h_1 h_2} h \paren {h_1 h_2}^{-1}$ $\displaystyle$ $=$ $\displaystyle h_1 h_2 h h_2^{-1} h_1^{-1}$ $\displaystyle$ $=$ $\displaystyle h_1 h' h_1^{-1}$ as $h' = h_2 h h_2^{-1} \in H_2$ $\displaystyle$ $=$ $\displaystyle h' h_1 h_1^{-1}$ $h_1 \in H_1$ and $h' \in H_2$ commute $\displaystyle$ $=$ $\displaystyle h' \in H_2$

Thus $a H_2 a^{-1} \subseteq H_2$ and $H_2$ is normal as well.

$\blacksquare$