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?


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.

  • $\begingroup$ 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 $\sigma$? $\endgroup$ – PerelMan Mar 10 at 21:37
  • $\begingroup$ 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. $\endgroup$ – jmerry Mar 10 at 21:43

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