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 a mapping defined by $\phi \left({h_1, h_2}\right) = h_1 h_2$.

If $\phi$ is an isomorphism, then both $H_1$ and $H_2$ are normal subgroups of $G$.

Proof

 * Let $\left({h_1, h_2}\right), \left({k_1, k_2}\right)$ be two elements of the group direct product of $H_1$ and $H_2$.


 * $\phi$ is an isomorphism, so by definition also a (group) homomorphism.

So, from Mapping from Cartesian Product Homomorphism iff Abelian, 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.

Thus by Internal Group Direct Product 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 $a h a^{-1}: h \in H_1$. Thus:

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 $a h a^{-1}: h \in H_2$. Thus:

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