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Prove that every finite group $G$ is isomorphic to a group of even permutations.

Let $G$ be a finite group . By Cayley's Theorem $G$ is isomorphic to a subgroup of $S_n$. Let $\tau $ be the required isomorphism under which $g\mapsto \tau(g)$.

If $\tau(g)$ is even we are done.If not we will have to turn $\tau(g)$ into an even permutation which I am failing to do.

Please help on how should I proceed.

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    $\begingroup$ Have you tried embedding $S_n$ into a group made up only of even permutations? $\endgroup$ – Steve D Mar 6 '16 at 3:58
  • $\begingroup$ It is Cayley's theorem, not Cauchy's. $\endgroup$ – Scientifica Mar 6 '16 at 6:16
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Claim: it is always true that $\;S_n\;$ can be embedded in $\;A_{n+2}\;$, because

$$\sigma\in S_n\implies \text{ map it as}\;\;\sigma \to\begin{cases}\sigma,&\text{if}\;\;\sigma\in A_n\\{}\\(n+1\;n+2)\sigma,&\text{if}\;\sigma\in S_n\setminus A_n\end{cases}$$

with $\;(n+1\;n+2)\;$ the transposition in $\;S_{n+2}\;$ interchanging $\;n+1\;$ and $\;n+2\;$

The above, together with Cayley's Theorem, gives the result.

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Define a homomorphism $S_n\to S_{2n}$ by sending a cycle $(a_1,a_2,\cdots,a_k)$ to the product of two disjoint cycles $(a_1,\cdots,a_k)(n+a_1,\cdots,n+a_k)$. This homomorphism is injective and its image lies entirely in $A_{2n}$. Thus you can construct an embedding $G\to S_n\to A_{2n}$.

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