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I had to show that $S_4$ has one subgroup of index $2$. Below, you'll find what I tried to do so


Of course, $A_4$ is a subgroup index $2$. To show that there is not another subgroup with the same index, we search all the normal subgroups of $S_4$. To do so, we need to know the cylcetypes that we could encounter:

  • (1,1,1,1)
  • (1,1,2)
  • (2,2)
  • (1,3)
  • (4)

If $\sigma$ belongs to some normal subgroup $N$, every element of the same cycletype as $\sigma$ must belong to $N$, because of the identity $\sigma (a_1, \cdots a_n) \sigma^{-1} \ = \ (\sigma(a_1), \cdots \sigma(a_n))$. Every cycle that can be written as a product of cycles of the same tape must be contained in $N$. Now we get the normal subgroups:

  • $\{e\}$
  • $S_n$
  • $\{(12)(34),(13)(24),(14)(23),e\}$
  • $A_n$
  • Maybe $A_n$ or else $S_n$.

For the last one I used that $(1234)(2134) = (143)$ That implies that this kind of cycles generates a normal subgroup that contains $A_4$. We see that the only normal subgroup containing $12$ elements is $A_4$, as we wanted to show.


Can you find flaws, or do you think this is a proof?

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Here's another approach, which works for any $n$: any subgroup of $S_n$ consists either of all even permutations, or has half even and half odd (multiplication by an odd permutation gives a bijection from evens to odds) - i.e., for any $H \le S_n$, $H \le A_n$ or $[H : A_n \cap H] = 2$.

Now if $[S_n : H] = 2$ and $H \ne A_n$, then $[H : A_n \cap H] = 2$, so $|A_n \cap H| = |H|/2 = |S_n|/4 = |A_n|/2$, so $[A_n : A_n \cap H] = 2$. But $A_n$ has no subgroup of index $2$, for any $n$.

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  • $\begingroup$ Is there a way to show that $A_n$ has no subgroup of index $2$ without using the simplicity of $A_n, n>4$? $\endgroup$ – Nishant Aug 3 '14 at 20:02
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    $\begingroup$ @Nishant: Well, simplicity is in my mind the easiest way. But for the case at hand, $n = 4$, the fact that $A_4$ has no subgroup of order $6$ is a well-known exercise, and can be done by hand $\endgroup$ – zcn Aug 3 '14 at 20:03
  • $\begingroup$ Oh right, I forgot that the OP was asking for the specific case of $n=1$. $\endgroup$ – Nishant Aug 3 '14 at 22:53
  • $\begingroup$ For those who would be interested, here is a proof of your last sentence. $\endgroup$ – Watson Oct 16 '16 at 12:40
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Here are some other suggestions.

Any subgroup of index $2$ is normal. The intersection of any two such subgroups is also normal.

Take $A_4$ as one of the two and $B$ as the hypothetical second. Then $A_4 \cap B$ is a normal subgroup of $S_4$ consisting of even permutations (because it is contained within $A_4$, and the intersection of normal subgroups is normal.

The only candidates are the trivial subgroup (use that a normal subgroup is a union of conjugacy classes), or the group $Z$ of order $4$ generated by the even permutations of order $2$.

Then use @zcn's observation that a subgroup has either all even permutations or half even/half odd to show that you can't have a subgroup of order $12$ with either $1$ or $4$ even elements.

Note also that you can use conjugacy class sizes to prove this directly

Trivial class - size 1

Type (12) - six elements

Type (12)(34) - three elements

Type (123) - eight elements

Type (1234) - six elements

If you want the unit element, you have to have the class of size three if the order of your normal subgroup is to be even ... and then you run out of options - only $A_4$ works.

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