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I was wondering, is it true that if $Alt_n$ is an alternating subgroup of $Sym_n$ for $n>3$, $Alt_{n-i}\leq Alt_n$ for all $i<n$?

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    $\begingroup$ What do you mean by $\le$ here? Does that mean that $Alt_{n-i} $ is a subgroup of $Alt_n$? $\endgroup$
    – MJD
    Mar 26 '15 at 16:20
  • $\begingroup$ Yes! I do mean that. $\endgroup$
    – insignia
    Mar 26 '15 at 16:21
  • $\begingroup$ More interesting is, that $S_n$ cannot be embedded into $A_{n+1}$, see here. $\endgroup$ Mar 26 '15 at 16:24
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The answer is no, in a strict sense. A permutation of $\{1,2,3\}$ is not a permutation of $\{1,2,3,4\}$. However, there is a natural bijection between the set of permutations of the former, and the set of permutations of the latter that fix $4$. Similarly, there is a natural bijection between $Alt_{n-i}$ and a subgroup of $Alt_n$, namely those permutations that fix $(n-i+1), (n-i+2),\ldots, n$.

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  • $\begingroup$ I guess that is what I was looking for then... I meant, is it true then that there exists a subgroup of $Alt_{n}$ that is isomorphic to $Alt_{n-i}$ for all $i<n$? $\endgroup$
    – insignia
    Mar 26 '15 at 16:25
  • $\begingroup$ Yes, a natural one, as described above. Just fix the other elements. $\endgroup$
    – vadim123
    Mar 26 '15 at 16:26
  • $\begingroup$ It is clear, but it should still be stated, that every permutation in $Alt_n$ in the range of the bijection is an even permutation: i.e. it really is in $Alt_n$ and not just in $Sym_n$. $\endgroup$ Mar 26 '15 at 16:30
  • $\begingroup$ In fact, there are several such isomorphisms besides the "obvious" one, as you can make $n$ distinct isomorphs of $S_{n-1}$ inside $S_n$, by choosing different elements to be "fixed". $\endgroup$ Mar 26 '15 at 19:20

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