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.