2

When $n\geq 5$ we can assume there is another one besides $A_n$, they are both normal, so the intersection is normal in $S_n$ and also in $A_n$, contradicting that $A_n$ is simple.

For $n=1$ it's trivial it is false, for $n=2$ it's trivial it is true, for $n=3$ the order of such a subgroup would be three, by cauchy it has a $3$-cycle, and if it does it is $\{(123),(132),e\}$

I'm just missing $S_4$, I think I did it before, but I would appreciate a clean and as slick as possible approach. (If you have to use advanced stuff to make it slicker I'd like it even more :p)

Thank you very much in advance

Regards.

Asinomás
  • 105,651
  • Re the first line, your proof is incomplete: if $H$ is another normal subgroup in $S_n$ of index two that is not trivial $H\cap A_n$ is normal in $A_n$. If $n>4$, this forces either $H\cap A_n=1$ or $H\cap A_n=A_n$. In the first case this cannot happen by cardinality considerations. It follows the second is true, $H\lhd A_n$, so $H=A_n$. – Pedro Jan 03 '15 at 01:42
  • I was actually implicitly using Timbuc's lemma which says the intersection of a sub of $S_n$ with $A_n$ can have either half the elements of the sub or all. – Asinomás Jan 03 '15 at 01:43

1 Answers1

2

Prove the following

Lemma: Let $\;H\le S_n\;$ , then either $\;H\le A_n\;$ or else

$$\;\left|H\cap A_n\right|=\frac{|H|}2$$

The above means that in the second possibility exactly half the elements of $\;H\;$ are even permutations.

Timbuc
  • 34,191
  • ok, I know that lemma, so then we assume exactly half are there, so we get $6$ even permutations, so then we get a subgroup of order $6$ of $A_4$. This doesn't exist, how would you prove that? – Asinomás Jan 03 '15 at 01:41
  • Is that how the proof goes? Am I right the contradiction is in that $A_4$ would have a sub of order $6$? – Asinomás Jan 03 '15 at 01:45
  • For $;A_4;$ yes, for all other $;n>4;$ not since $;S_3;$ can be embedded in $;A_5;$ . In fact, $;S_n;$ can be embedded in $;A_{n+2};$ – Timbuc Jan 03 '15 at 01:46
  • yeah, I meant $A_4$, all the other cases are covered. How would you prove $A_4$ has no sub of order $6$ as slick as possible? – Asinomás Jan 03 '15 at 01:47
  • And I didn't meant about a subgroup of order six or whatever, but about simplicity of $;A_n;,;;n\ge 5;$ . – Timbuc Jan 03 '15 at 01:47
  • yeah, I answered all orders except $4$ in my original post, I just need $S_4$. – Asinomás Jan 03 '15 at 01:48
  • @JorgeFernández, perhaps showing that a subgroup generated by an element of order two and one of order three in $;A_4;$ is, after some calculations, already too big (because, clearly, there is no element of order six in $;A_4;$) – Timbuc Jan 03 '15 at 01:48
  • @JorgeFernández Read theorem 1 here: http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/A4noindex2.pdf It's as easy and elementary a proof as you can expect, imo. – Timbuc Jan 03 '15 at 01:50
  • That's perfect. That's really all that was missing, could you just write that in your solution? A normal subgroup of $S_4$ would have all the odd permutations, so it would have all the elements of $A_4$, and no others. – Asinomás Jan 03 '15 at 01:56
  • In fact that solution is great, you don't even need normality, you just need that half of the elements of $S_n$ are of odd order, which is really simple. – Asinomás Jan 03 '15 at 01:57
  • @JorgeFernández Well, you do need normality of the group $;N;$ in that theorem 1 for the quotient group $;G/N;$ to exist. – Timbuc Jan 03 '15 at 02:08
  • yeah, you need the normality of $N$, but not of $A_n$ – Asinomás Jan 03 '15 at 02:09
  • I'll accept your answer, although what I needed is what is in the link, what is currently in the answer text I didn't help that much. But that link is awesome, thanks. – Asinomás Jan 03 '15 at 02:10