Internal Group Direct Product Isomorphism

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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:

\(\displaystyle a h a^{-1}\) \(=\) \(\displaystyle \paren {h_1 h_2} h \paren {h_1 h_2}^{-1}\)
\(\displaystyle \) \(=\) \(\displaystyle h_1 h_2 h h_2^{-1} h_1^{-1}\)
\(\displaystyle \) \(=\) \(\displaystyle h_1 h h_2 h_2^{-1} h_1^{-1}\) $h \in H_1$ 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.

$\blacksquare$