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We have two groups $X$ and $Y$ of prime order, when is semidirect product defined from different homomorphism to the automorphism group same?

I have read When are two semidirect products isomorphic? but it may be easier to determine when the order of the groups are primes.

My try I don't see why can they be isomorphic. We have two distinct homomorphisms and therefore $1$ maps to distinct automorphisms of $H$. For semidirect products to be same, image of $1$ should be the same automorphism, by the way we define product in the external semidirect product.

Please help.

Martund
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1 Answers1

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Here's an example that might help: Let $X$ have order $7$ and $Y$ have order $3$. These groups are cyclic; write $x$ for a generator of $X$ and $y$ for a generator of $Y$. Let $\phi\colon X \to X$ and $\psi\colon X \to X$ be the automorphisms $\phi(x) = x^2$, and $\psi(x) = x^4$.

Notice that $\phi^3(x) = x^{2^3} = x^8 = x$, and $\psi^3(x) = x^{4^3} = x^{64} = x$, so as automorphisms of $X$, $\phi$ and $\psi$ each have order three.

I claim that $X\rtimes_\phi Y \cong X\rtimes_\psi Y$. Indeed, (abusing notation slightly) $X\rtimes_\phi Y$ and $X\rtimes_\psi Y$ are generated by $x$ and $y$, so to specify a homomorphism, I just need to tell you where $x$ and $y$ go, and then check that the map I've written down indeed defines a homomorphism. I claim that the map $x \mapsto x$, $y \mapsto y^{-1}$ is such a homomorphism and moreover that it is an isomorphism. I'll leave it to you to check this.

So if the answer has to be "sometimes" and not "never," maybe we should reconsider what should be true based on this example. Aut$(X)$ is generated by $x \mapsto x^3$. There is an automorphism of Aut$(X)$ that sends the automorphism $x \mapsto x^3$ to the automorphism $x \mapsto x^5$. Under this automorphism of Aut$(X)$, $\phi$ is mapped to $\psi$.

So some questions to investigate: if we have $\phi$ and $\psi$ in Aut$(X)$, will it be the case that $X\rtimes_\phi Y\cong X\rtimes_\psi Y$ whenever there is some automorphism $f\colon$ Aut$(X) \to $ Aut$(X)$ such that $f(\phi) = \psi$? Is this the only condition?

Robbie Lyman
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  • Where I am confused is the following. $Y$ is isomorphic to $\mathbb Z_{|Y|}$ and $Aut(X)$ is isomorphic to $\mathbb Z_{|X|-1}$, taking a homomorphism here, semidirect products are same iff $\Phi_1(n)(m)=\Phi_2(n)(m)\forall n\in\mathbb Z_{|Y|} ,m\in\mathbb Z_{|X|-1}$ . Now, by property of homomorphisms, this happens iff $\Phi_1(1)(1)=\Phi_2(1)(1)$, which cannot happen because $\Phi_1$ and $\Phi_2$ are distinct homomorphisms. Please clarify. – Martund Dec 23 '19 at 02:07
  • Here is the false statement: "semidirect products are isomorphic if and only if $\Phi_1(n)(m) = \Phi_2(n)(m)$ for all $n \in \mathbb Z_{|Y|}$ and $m\in\mathbb Z_{|X|-1}$." I have given an example above that proves that the "only if" direction is false. – Robbie Lyman Dec 23 '19 at 02:39
  • That is a doubt. Without this, how is multiplication same in semidirect product by two homomorphisms, because $(x_1,y_1).(x_2,y_2)=(x_1+\Phi(y_1)(x_2),y_1+y_2)$ – Martund Dec 23 '19 at 02:42
  • Ahh. You seem to be using two groups are "the same" to mean "identical". I am using "the same" to mean "there exists a group isomorphism from one to the other." – Robbie Lyman Dec 23 '19 at 02:44
  • I am not getting what $\Phi:Y\to Aut(X)$ are you taking. – Martund Dec 23 '19 at 02:44
  • I guess if we identify $Aut(X)$ with $\mathbb Z_6$, then $\phi\colon Y \to Aut(X)$ is $1 \mapsto 2$. $\psi\colon Y \to Aut(X)$ is $1 \mapsto 4$. – Robbie Lyman Dec 23 '19 at 02:46