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This question already has an answer here:

I want to show that $S_n \cong A_n \rtimes C_2$.

Take a transposition $\tau \notin A_n$. Then it is clear that

$$\langle \tau\rangle \cap A_n = 1$$ $$A_n \tau = S_n$$ $$A_n \unlhd S_n$$

and thus $$S_n \cong A_n \rtimes_\phi \langle \tau \rangle$$

with $$\phi: \langle \tau \rangle \to Aut(A_n): \mathcal{\tau} \mapsto (A_n \to A_n: \sigma \mapsto \tau^{-1} \sigma \tau = \tau \sigma \tau), 1 \mapsto 1$$

I'm now trying to see what $\tau^{-1} \sigma \tau$ looks like, as I would like to write:

$$A_n \rtimes_\phi \langle \tau \rangle \cong A_n \rtimes_\psi C_2$$

for some homomorphism $\psi: C_2 \to Aut(A_n)$.

How can I proceed?

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marked as duplicate by Arnaud D., Dietrich Burde group-theory Apr 19 at 11:22

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ As for the last bit, recall that for any element $\rho$ of $S_n$, if $S_n$ acts on $\{1,\dots,n\}$ on the right, then $\rho^{-1}(n_1\dots n_k)\rho = (n_1\rho\dots n_k\rho)$. So decomposing elements of $A_n$ as products of disjoint cycles should give us some information. $\endgroup$ – Rylee Lyman Apr 19 at 11:05
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Being of index $2$, $A_n$ is normal in $S_n$ and from the inclusion we get the short exact sequence $$\tag1 1\to A_n\to S_n\to C_2\to 1.$$ As you said, we can pick a transposition $\tau\in S_n$, so $\tau^2=1$ and $\tau\notin A_n$. Then sending the nontrivial element of $C_2$ to $\tau$ makes $(1)$ split, which means that $S_n$ is a semi-direct product of $A_n$ by $C_2$.

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  • $\begingroup$ Thanks for your answer! I have a follow up question. Is it possible to construct $S_n $ using this semidirect product? I.e., given the group structures of $C_2$ and $A_n$, can I find the entire group structure of $S_n$? I think my question boils down to: can we find the explicit homomorphism $\phi$ such that $S_n \cong A_n \rtimes_\phi C_2$, without it depending on anything in $S_n$. $\endgroup$ – user661541 Apr 19 at 11:04
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    $\begingroup$ See the comments to this question, answering your question. $\endgroup$ – Dietrich Burde Apr 19 at 11:12
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    $\begingroup$ @user661541 Any involutory automorphism of $A_n$ will do - but I see no direct way of constructing such an automorphism from "within" $A_n$, i.e., without actually referencing $S_n$. Only exception: $A_3\cong C_3$ is abelian, hence inversion is such an automorphism and can be used to construct $S_3$ from it. $\endgroup$ – Hagen von Eitzen Apr 19 at 11:19

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