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Let $n\geq5$, $G$ be a subgroup of $S_n$ s.t. $G$ has no subgroups of index 2 (G is also simple) and there is an injection (morphism) from $G$ to $S_n$.

Is this enough to say that $G$ lay in $A_n$ ? as $A_n$ is simple and has no subgroup of index 2.

Is there any use of the 2nd isomorphism theorem to prove this? i'm working on a group of order 2016 and for n=8 ($o(G) \leq o(A_{8})$)

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    $\begingroup$ You don't need to assume that $G$ is simple, because if $G$ is not contained in $A_n$, then $G \cap A_n$ is a subgroup of index $2$ in $G$. $\endgroup$ – Derek Holt Dec 26 '15 at 17:51
  • $\begingroup$ Well by the use of the 2nd isomorphim theorem, i can see that $G \cap A_{n}$ is normal in $G$.. but i don't understand why it is a subgroup of index $2$ in $G$ $\endgroup$ – LeLoupSolitaire Dec 26 '15 at 18:04
  • $\begingroup$ Because $|S_n:A_n|=2$. $\endgroup$ – Derek Holt Dec 26 '15 at 19:16
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    $\begingroup$ Aaah i see, we use the fact the $\left| G : G \cap A_{n} \right| \leq \left| S_{n} : A_{n} \right| = 2 $, and as $G$ is not contained in $A_n$ , $G \cap A_{n} \neq G$ and the index can't be equal to $1$, thank you very much Derek, it seems more clear now =) $\endgroup$ – LeLoupSolitaire Dec 26 '15 at 20:01
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If $G$ is simple non commutative, $[G,G]=G\subset [S_n,S_n]=A_n$ done.

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  • $\begingroup$ Interesting fact, but what if we do not know that $G$ it is simple, but just that it does not have a subgroup of index $2$ ? does this still true ? $\endgroup$ – LeLoupSolitaire Dec 26 '15 at 18:16
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We do not need whether $A_n$ is simple or not; we need to know that its index in $S_n$ is $2$, which is true for all $n\geq 1$. You may ignore also the hypothesis that $G$ is simple.

Suppose $G$ is not in $A_n$ (but it is in $S_n$ by htpothesis). Then $G.A_n=S_n$ and hence $G\cap A_n$ is subgroup of index $2$ in $G$ (use the formula for $|HK|=|H|.|K|/|H\cap K|$ or any other way).

This contradicts another hypothesis. Hence $G$ must be in $A_n$.

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