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.
Thus by Subgroup Product is Internal Group Direct Product iff 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:
\(\ds a h a^{-1}\) | \(=\) | \(\ds \paren {h_1 h_2} h \paren {h_1 h_2}^{-1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds h_1 h_2 h h_2^{-1} h_1^{-1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds h_1 h h_2 h_2^{-1} h_1^{-1}\) | $h \in H_1$ and $h_2 \in H_2$ commute | |||||||||||
\(\ds \) | \(=\) | \(\ds h_1 h h_1^{-1} \in H_1\) |
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:
\(\ds a h a^{-1}\) | \(=\) | \(\ds \paren {h_1 h_2} h \paren {h_1 h_2}^{-1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds h_1 h_2 h h_2^{-1} h_1^{-1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds h_1 h' h_1^{-1}\) | as $h' = h_2 h h_2^{-1} \in H_2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds h' h_1 h_1^{-1}\) | $h_1 \in H_1$ and $h' \in H_2$ commute | |||||||||||
\(\ds \) | \(=\) | \(\ds h' \in H_2\) |
Thus $a H_2 a^{-1} \subseteq H_2$ and $H_2$ is normal as well.
$\blacksquare$