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I'm solving the following problem:

If $f:S_n\rightarrow S_n$ is a group homomorphism, prove that $f(A_n)\subseteq A_n.$ (Here, $S_n$ is a symmetric group of degree $n$, and $A_n$ is an alternating group of degree $n.$)

For $n=2,$ it is trivial. Let $n\geq3.$ First we show that for any $3$-cycle $(abc)\in S_n,$ its image $f((abc))$ is even. Suppose, on the contrary, that $f((abc))$ is odd. Since $(abc)^3=(1),$ $f((abc))^3=f((abc)^3)=f((1))=(1)$. (Note that $f$ is a homomorphism). Thus, $f((abc))^3=(1).$ However, $(1)$ is $even$ and since we assumed that $f((abc))$ is odd, $f((abc))^3$ is odd. This is a contradiction! Thus $f((abc))$ is even. As every element $\sigma$ of $A_n$ (that is, all even permutations) is a product of $3$-cycles (Link), we may write $\sigma = (a_1b_1c_1)\cdots(a_nb_nc_n).$ Then, $f(\sigma)=f((a_1b_1c_1)\cdots(a_nb_nc_n))=f((a_1b_1c_1))\cdots f((a_nb_nc_n)).$ As each $f((a_1b_1c_1)),\dots,f((a_nb_nc_n))$ is even, $f(\sigma)$ is also even. It follows that $f(A_n)\subseteq A_n$!

Is my argument correct?

user26857
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Kim
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    Yes; but it would be simpler to note that $f(acb) = (f(abc))^{-1}$, and $f(abc) = f((acb)^2) = (f(acb))^2$ is the square of an element of $S_n$, and hence must lie in $A_n$. – Arturo Magidin Apr 21 '20 at 18:46
  • @ArturoMagidin Oh, I haven't thought about it that way. Thank you! – Kim Apr 21 '20 at 19:01

1 Answers1

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Yeah, it's correct. In fact, once you have shown that for $n \geq 5$, the only proper normal subgroup of $S_n$ is $A_n$, then since $\ker f$ is normal, if $\ker f = S_n$, $f$ is the zero-mapping, if $\ker f =A_n$ then $f(A_n)=(1)$ and if $\ker f = (1)$, then $f$ is bijective and thus sends normal subgroups to normal subgroups and in that case, $f(A_n)=A_n$.

Edit: This is not true for $n=4$ because $S_4/V_4 \cong S_3$ (where $V_4 = \left\langle e, (12)(34), (23)(14), (13)(24) \right\rangle$ is the Klein group), which can be embedded in $S_4$ without containing the whole $A_4$.

user26857
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Aitor Iribar Lopez
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