# Conjugate images induce isomorphic semi-direct products?

See here for the question context.

Let $$H_5 = \Bbb Z_5\times \Bbb Z_5$$. I am trying to prove that for all non-trivial homomorphism $$\varphi:\Bbb Z_3\to \operatorname{Aut}(H_5),$$ the resulting semi-direct products $$\Bbb Z_3\ltimes_{\varphi} H_5$$ are all isomorphic. I know $$\left|\operatorname{Aut}(H_5)\right| = 2^5\cdot 3\cdot 5$$ so the non-trivial image of $$\varphi(\Bbb Z_3)$$ are all conjugates to each other as Sylow $$3$$-subgroups in $$\operatorname{Aut}(H_5)$$. If we take non-trivial homomorphisms $$\phi,\varphi:\Bbb Z_3\to\operatorname{Aut}(H_5)$$ I am guessing if we write $$\varphi(\Bbb Z_3) = \gamma\phi(\Bbb Z_3)\gamma^{-1}$$ for some $$\gamma\in\operatorname{Aut}(H_5)$$ then $$\Bbb Z_3\ltimes_{\phi} H_5\to \Bbb Z_3\ltimes_{\varphi} H_5, (x,y)\mapsto (x, \gamma(y))$$ should be an isomorphism.

But I am having trouble to prove that it is a homomorphism. What I will need is $$(x_1x_2,\gamma(\phi_{x_2}(y_1)y_2)) = (x_1x_2,\varphi_{x_2}(\gamma(y_1))\gamma(y_2)),$$ or that $$\varphi=\gamma\circ\phi\circ\gamma^{-1}$$

This requires stronger condition than $$\varphi(\Bbb Z_3) = \gamma\phi(\Bbb Z_3)\gamma^{-1}$$ which we don't have. How should I proceed?

• I think you just didn't do the computations properly. Commented Oct 5, 2023 at 20:03

I think you just didn't do the calculations properly.

Let $$C_n$$ denote the cyclic group of order $$n$$ (so I don't have to type $$\mathbb{Z}_5$$ which is harder to $$\LaTeX$$ and can also be confused with the $$5$$-adic integers. Let $$H=C_5\times C_5$$.

Let $$\varphi,\theta\colon\mathbb{C}_3\to\mathsf{GL}_2(5)$$ be two nontrivial homomorphisms. We want to show that $$H\rtimes_{\varphi}C_3\cong H\rtimes_{\theta} C_3$$. This will require some specific facts about your specific orders, since the result is not true in general.

An element of order $$3$$ in $$\mathsf{GL}_2(5)$$ satisfies $$X^3-1=(X-1)(X^2+X+1)$$. Since both factors are irreducible in $$\mathbb{F}_5$$, it means that the minimal polynomial is either $$X^2+X+1$$ or $$X-1$$; in the latter case it is the identity, which we are excluding, so the minimal polynomial is $$X^2+X+1$$. That means that matrix for this element (with, say, the standard basis for $$C_5\times C_5$$) must be conjugate to the companion matrix of $$X^2+X+1$$. Thus, any two elements of order $$3$$ in $$\mathsf{GL}_2(5)$$ are conjugate in $$\mathsf{GL}_2(5)$$. That is, there exists an automorphism $$\psi\colon H\to H$$ such that $$\theta = \psi\varphi\psi^{-1}$$.

If we denote the automorphism $$\varphi(y)$$ by $$\varphi_y$$, and the automorphism $$\theta(y)$$ by $$\theta_y$$, we have that for each $$y\in Y$$, $$\theta_y = \psi\varphi_y\psi^{-1}$$.

(Note: I write my semidirect products backwards from how you write them: the first coordinate lies in the normal subgroup, the second in the "acting" group. So the product in $$A\rtimes_g B$$ is $$(a_1,b_1)(a_2,b_2) = (a_1g_{b_1}(a_2), b_1b_2)$$. You can translate this to your notation..)

Define $$f\colon H\rtimes_{\varphi}C_3 \to H\rtimes_{\theta} C_3$$ as follows: $$f(x,y) = (\psi(x), y).$$ We have \begin{align*} f(x,y)f(r,s) &= (\psi(x),y)(\psi(r),s)\\ &= \Bigl(\psi(x)\theta_y\bigl(\psi(r)\bigr),ys\Bigr)\\ &= \Biggl(\psi(x)\biggl(\psi\Bigl(\varphi_y\bigl(\psi^{-1}(\psi(r)\bigr)\Bigr)\biggr),ys\Biggr)\\ &= \Bigl(\psi(x)\psi\bigl(\varphi_y(r)\bigr),ys\Bigr)\\ &= \Bigl(\psi\bigl(x\varphi_y(r)\bigr),ys\Bigr)\\ &= f\Bigl(x\varphi_y(r),ys\Bigr). \end{align*}

Now note that in $$H\rtimes_{\varphi}C_3$$, we have $$(x,y)(r,s) = (x\varphi_y(r),ys).$$ Therefore, we have that $$f\Bigl((x,y)(r,s)\Bigr) = f(x,y)(r,s).$$ The map $$g\colon H\rtimes_{\theta}C_3 \to H\rtimes_{\varphi}C_3$$ given by $$g(x,y) = (\psi^{-1}(x),y)$$ is a homomorphism by the same argument, and it is clearly the inverse of $$f$$, so $$f$$ is an isomorphism, as desired.

The calculation shows that in general, if $$\varphi,\theta\colon K\to \mathrm{Aut}(N)$$, and there exists $$\psi\in\mathrm{Aut}(N)$$ such that $$\varphi=\psi\theta\psi^{-1}$$, then $$N\rtimes_{\varphi}K\cong N\rtimes_{\theta}K$$; the result is not restricted to finite groups.

• The linear algebra part is what I am missing. Thanks! Commented Oct 5, 2023 at 22:18
• @user108580 That just proves the existence of $\psi$; but that is not what you are asking in the post... the post is asking why conjugate actions yield isomorphic semidirect products. That comment makes no sense. Commented Oct 5, 2023 at 22:31
• I believe your $\psi$ is exactly the $\gamma$ that I am asking for that conjugates with $\phi$ and $\varphi$. This is the only part I am missing as written in my question. Commented Oct 5, 2023 at 22:35
• @user108580 No... your post states that you are having trouble proving that the map I denote as $f$ is a homomorphism. That map already assumes the existence of your $\gamma$. If that was your question, then your post does not ask that question. Your post is asking for a proof that $f$ is a homomorphism, not for a proof that $\gamma$ exists. Commented Oct 5, 2023 at 22:37
• @user108580: I mean, look at the title of your post! It says "conjugate images induce isomorphic semidirect products?" Not "are any two of these actions conjugate?" Commented Oct 5, 2023 at 22:38