# isomorphism between semi-direct products

given two groups H and K, and two morphisms $$φ$$ and $$φ'$$ from K to $$Aut(H)$$.

given $$σ \in Aut(K)$$ such that $$φ' = φ ◦ σ$$, prove that $$H \rtimes_φ K \cong H \rtimes_{φ'} K$$.

I found this simple isomorphism $$f: H \rtimes_φ K \to H \rtimes_{φ'} K$$ defined by $$f(h,k) = (h, σ(k))$$ which is injective and surjective by construction. Is that correct or there is a more subtle answer?

We just have to check it against the multiplication rule of the two groups, to see whether it's a homomorphism. The product $$(h_1,k_1)\cdot (h_2,k_2)$$ in the first group is $$(h_1\varphi(k_1)(h_2),k_1k_2)$$. In the second group, we have the product of $$(h_1,\sigma(k_1))$$ and $$(h_2,\sigma(k_2))$$, which is $$f(h_1,k_1)\cdot f(h_2,k_2) = (h_1,\sigma(k_1))\cdot (h_2,\sigma(k_2)) = (h_1\varphi'(\sigma(k_1))(h_2),\sigma(k_1)\sigma(k_2))$$ $$(h_1\varphi'(\sigma(k_1))(h_2),\sigma(k_1)\sigma(k_2)) \stackrel{?}{=} (h_1\varphi(k_1)(h_2),\sigma(k_1k_2)) = f((h_1,k_1)\cdot (h_2,k_2))$$ That works in the second coordinate, but not in the first; we would need $$\varphi'\circ\sigma=\varphi$$, and we instead have $$\varphi'=\varphi\circ\sigma$$. Indeed, this isn't correct.
How do we repair it? Simple - use $$f(h,k)=(h,\sigma^{-1}(k))$$ instead. Or, equivalently, turn the arrow around.
• Thanks! in the second formulae to check. do you mean $(h_1\varphi'(\sigma(k_1))(h_2),\sigma(k_1)\sigma(k_2)) \stackrel{?}{=} (h_1\varphi(k_1)(h_2),k_1k_2)$? because in the first group there should be no mention of $\sigma$? – PerelMan Mar 10 at 21:37
• The $\stackrel{?}{=}$ line is comparing the product $f(h_1,k_1)\cdot f(h_2,k_2)$ with $f((h_1,k_1)\cdot (h_2,k_2))$. Actually, I'll edit to make that clearer; I hadn't noticed that $f$ has a name while writing that part, and was working around it a bit awkwardly. – jmerry Mar 10 at 21:43