# Show that $S_n \cong A_n \rtimes C_2$ [duplicate]

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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?

## marked as duplicate by Arnaud D., Dietrich Burde group-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 19 at 11:22

• 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. – Rylee Lyman Apr 19 at 11:05
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 $$C_2$$.
• 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 $S_n$. – user661541 Apr 19 at 11:04
• @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. – Hagen von Eitzen Apr 19 at 11:19