Semidirect Product with Trivial Action is Direct Product
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Theorem
Let $H$ and $N$ be groups.
Let $\Aut N$ denote the automorphism group of $N$.
Let $\phi: H \to \Aut N$ be defined as:
- $\forall h \in H: \map \phi h = I_N$ for all $h \in H$
where $I_N$ denotes the identity mapping on $N$.
Let $N \rtimes_\phi H$ be the corresponding semidirect product.
Then $N \rtimes_\phi H$ is the direct product of $N$ and $H$.
Proof
Pick arbitrary $\tuple {n_1, h_1}, \tuple {n_2, h_2} \in N \rtimes_\phi H$.
\(\ds \tuple {n_1, h_1} \tuple {n_2, h_2}\) | \(=\) | \(\ds \tuple {n_1 \cdot \map \phi {h_1} \paren {n_2}, h_1 h_2}\) | Definition of Semidirect Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \tuple {n_1 \cdot \map {I_N} {n_2}, h_1 h_2}\) | Definition of $\phi$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \tuple {n_1 n_2, h_1 h_2}\) |
which meets the definition of direct product.
$\blacksquare$