The title is the question. Is $A_n$ characteristic in $S_n$?

If $\phi \in \operatorname{Aut}(S_n)$, Then $[S_n : \phi(A_n)]$ (The index of $\phi(A_n)$) is 2. Maybe the only subgroup of $S_n$ of index 2 is $A_n$?

Thanks in advance.

  • $\begingroup$ Homework question? $\endgroup$ – rogerl Apr 18 '12 at 1:18
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    $\begingroup$ $A_n$ is commutator of $S_n$. $\endgroup$ – user641 Apr 18 '12 at 1:32
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    $\begingroup$ In fact, since $A_n=[S_n,S_n]$, $A_n$ is fully invariant: if $f\colon S_n\to S_n$ is any endomorphism, then $f(A_n)\subseteq A_n$. $\endgroup$ – Arturo Magidin Apr 18 '12 at 2:41

Definition: We say that $H \leqslant G$ is a characteristic subgroup of $G$ if every isomorphism fixes $H$. That is, $$\phi(H) \subseteq H\;\; \mbox{for every isomorphism}\;\; \phi:G\to G $$

Yes, $A_n$ is characteristic in $S_n$.


  1. That $A_n$ is a unique subgroup of index $2$ tells us that $A_n$ must be sent to itself by automorphism.

    • To see that $A_n$ is a unique of subgroup of index $2$, for the homomorphism, $\theta$ from $S_n$ to $C_2$, note that every element of odd order is in the kernel, in particular, all $3-$cycles. Since all three cycles generate $A_n$, we must have that $A_n$ is in the kernel, but $|A_n|=|\ker \theta |$ which means $A_n=\ker \theta$.
  2. $[S_n,S_n]=A_n$. It is easy to see that commutator subgroup of $G$ is characteristic in $G$. (Hint: Commutator subgroup is generated by commutators; what is the image of a commutator under an automorphism?)

    • To see $[S_n,S_n]=A_n$, note that, $S_n /A_n \simeq C_2 $ which is abelian, we should have that $[S_n,S_n]\subseteq A_n$. Alternatively, every three cycle is a commutator of two transpositions, viz, $[(a b), (a c)]=(a b c)$. This means every three cycle and hence $A_n$ is in the commutator subgroup which proves the claim.
  3. Automorphisms preserve parity of the elements.

    • Inner automorphisms and only inner automorphisms preserve cycle type; $S_6$ has an exceptional outer automorphism which takes transpositions to product of three transpositions and hence parity is preserved. The proof of this fact is discussed in Rotman's book, Introduction to the theory of Groups.

Note: The only proper normal subgroup in $S_n$ for $n \geq 5$ is $A_n$. Thus, these groups give examples where every normal subgroup is characteristic.

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    $\begingroup$ -1; only inner automorphisms preserve cycle type. $S_6$ has an exceptional outer automorphism which doesn't: en.wikipedia.org/wiki/… $\endgroup$ – Qiaochu Yuan Apr 18 '12 at 1:35
  • $\begingroup$ One should also prove that all automorphisms of $S_n$, $n \neq 2, 6$ are inner. Rotman's book seems to say a lot about both points. The outer automorphisms of $S_6$ send each transposition to a product of three transpositions, so parity is preserved and the statement is still true. $\endgroup$ – Dylan Moreland Apr 18 '12 at 1:37
  • $\begingroup$ @QiaochuYuan I missed this point totally: that $A_n$ is the unique subgroup of order $n!/2$ should complete the proof as well. (I was also misled: in that I wanted to make a statement for parity and not cycle type.) $\endgroup$ – user21436 Apr 18 '12 at 1:41
  • $\begingroup$ @Dylan Thank you for reminding me that I wanted make a statement about parity and not cycle types. $\endgroup$ – user21436 Apr 18 '12 at 1:44
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    $\begingroup$ @Kannappan: what you mean by "parity" is worth explaining. From one point of view it is just the map to the abelianization $G \to G/[G, G]$ and so the third argument is actually the same as the second. $\endgroup$ – Qiaochu Yuan Apr 18 '12 at 2:56

Let $H$ be a subgroup of $S_n$. Then either $H \subseteq A_n$ or $[H:H\cap A_n]=2$ (that is either $H$ is all even or half-even and half-odd). This fact can be proven by noticing that left multiplication by an odd permutation (if there is one in $H$) sends evens to odds and odds to evens bijectively (so there must be an equal number of both evens and odds if there are any odd elements to begin with).

Therefore, the only subgroup of index $2$ in $S_n$ is $A_n$ [If $H$ is all even and index $2$, it must be all of $A_n$. If $H$ is half-even and half-odd, then $A_n$ has a normal subgroup: $H \cap A_n$ of index 2 in $A_n$, but no such subgroup exists by inspection for $n=1,2,3,4$ and simplicity of $A_n$ for $n \geq 5$]. Thus $A_n$ is characteristic (being the unique subgroup of $S_n$ of order $n!/2$ it must be sent to itself by any automorphism).

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    $\begingroup$ Would it be better to say that "half-even and half odd" means (calling $H$ the subgroup) $(H : H \cap A_n) = 2$? $\endgroup$ – Dylan Moreland Apr 18 '12 at 1:44
  • $\begingroup$ Yes. That's what I meant.... :P $\endgroup$ – Bill Cook Apr 18 '12 at 1:50

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