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I want to prove $A_n$ has no subgroups of index 2.

I know that if there exists such a subgroup $H$ then $\vert H \vert = \frac{n!}{4}$ and that $\vert \frac{A_n}{H} \vert = 2$ but am stuck there. I have tried using the proof that $A_4$ has no subgroup of order 6 to get some ideas but am still stuck. Sorry I don't have much else to add at this point. Thanks a bunch.

user26857
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J. Kennedy
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    Are you allowed to use the fact that the commutator subgroup of $A_n$ is the group itself if $n \ge 5$? – Nicky Hekster Oct 28 '15 at 15:26
  • @Morgan, you do not need that - if $H$ is a subgroup of index $2$, then $H$ is normal and $A_n/H$ is abelian (of order $2$), whence$A_n' \subseteq H$. – Nicky Hekster Oct 28 '15 at 15:53

3 Answers3

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This elementary argument works for all $n$ at once and does not need simplicity of the alternating group: Let $N$ be of index $2$ in $A_n$ and let $x$ be a $3$-cycle. We have $xN=x^4N=(xN)^4=N$, where the latter holds since the factor group is of order $2$. We derive that $N$ contains all $3$-cycles. But it is a well known (and easy to show) fact that the alternating group is generated by $3$-cycles.

Of course we implictly used that a subgroup of index $2$ is always normal.

MooS
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  • Should $x$ be a 3-cycle in $A_n$ but not in $N$? – J. Kennedy Oct 28 '15 at 18:06
  • $x$ is any $3$-cycle (of course a $3$-cycle is an even permutation). My argument shows that $x$ is then automatically contained in $N$. But this is a contradiction, since a subgroup of $A_n$ containing all $3$-cycles is whole of $A_n$, hence of index $1$. – MooS Oct 28 '15 at 20:26
  • I see, thank you for the clarification. I continued working on this after I posted and came up with another possible proof. If I wanted to have my proof looked over, should I post it here or begin a new question thread? – J. Kennedy Oct 29 '15 at 00:27
  • Feel free to post as an answer in this thread. – MooS Oct 29 '15 at 04:57
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As has been stated in the comments, if you know that $A_n$ is simple for $n\geq 5$, and subgroups of index 2 are normal, you are done with that case.

For the $n=4$ case, you do need to do a bit more work. Your idea to show that $A_n$ has no subgroup of order 6 is correct. Well, a subgroup of order $6$ in $A_4$ is either cyclic, or isomorphic to $S_3$. Since $A_4$ has no element of order 6, we can exclude the case of a cyclic subgroup and any subgroup of $A_4$ of order 6 must be isomorphic to $S_3$.

To see that this can't happen, suppose $$\phi:S_3\to A_4$$ is an injective homomorphism and use the fact that $S_3$ is generated by the simple transpositions $(12),(23)$. The image of these elements must be elements of order 2 in $A_4$ which are disjoint 2-cycles. Any two of these generate a subgroup of order 4 in $A_4$. Hence, no such homomorphism exists.

David Hill
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    Another approach for the remaining case $n=4$: A subgroup of index 2 contains all squares. But $A_4$ has eight $3$-cycles, which happen to be squares, since we have $x=x^4$ for a $3$-cycle. – MooS Oct 28 '15 at 16:24
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Hint: What do we know about a subgroup of $G$ whose index is the smallest prime dividing $|G|$?

Espen Nielsen
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