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Q1 Prove that every simple subgroup of $S_4$ is abelian.

Q2 Using the above result, show that if $G$ is a nonabelian simple group then every proper subgroup of $G$ has index at least $5$.

My attempt

For Q1:

$|S_4| = 4\cdot3\cdot2 = 2^3 \cdot 3$.

One can exhibit all possible orders of subgroups of $S_4$ and eliminate subgroups according to the requirement that the subgroup is simple.

For example, any subgroup of order $2\cdot 3$ is not simple so needs not to be considered.

Then it might be possible to show that all remaining subgroups are abelian using results like a group of order $p^2$, where $p$ is a prime, is abelian.

For Q2

I think Cayley's Theorem is involved and I should probably consider the quotient of $G$ on a proper subgroup of $G$. But since $G$ is simple, the quotient is not a group. And I don't really know where to go from here.

My question

For Q1, I'm not sure that the above approach would work and even if it does, I feel like it is too cumbersome and there might be a more principled and more concise way.

Any help would be greatly appreciated.

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  • $\begingroup$ Please ask one question at a time. $\endgroup$ – Shaun May 5 '19 at 18:15
  • $\begingroup$ @Shaun Given that the second question relies heavily on the result of the first question, I do not see the need to break them into two separate posts. $\endgroup$ – msd15213 May 5 '19 at 18:17
  • $\begingroup$ Fair enough :) ${}$ $\endgroup$ – Shaun May 5 '19 at 18:18
  • $\begingroup$ @Shaun thanks ;) $\endgroup$ – msd15213 May 5 '19 at 18:19
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For the first question, every simple non-abelian group has order at least $60$, and hence cannot be a subgroup of $S_4$, which has only $24$ elements.

Reference: $G$ is non abelian simple group of order $<100$ then $G\cong A_5$

For the second question, if $G$ is a simple group with a subgroup of index $n>1$, then $G$ injects into the symmetric group $S_n$.

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    $\begingroup$ I think the result you are using to answer the first question is more advanced than the question itself. $\endgroup$ – Derek Holt May 5 '19 at 15:16
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    $\begingroup$ @DerekHolt True, but on the other hand we only need to look for the cases of divisors of $24$, which are included in the other answer, too. So we do not need the "advanced cases". The case $n=24$ is discussed in detail, too. $\endgroup$ – Dietrich Burde May 5 '19 at 15:17
  • $\begingroup$ @DietrichBurde I looked at the other question you referred to and it's quite some work lol. But thanks I think I will spend some time on understanding that question. $\endgroup$ – msd15213 May 5 '19 at 15:36
  • $\begingroup$ @DietrichBurde About your answer to my second question, though. Could you please give some details on the specific map that we can construct from $G$ to $S_n$? I am trouble finding one. Thanks. $\endgroup$ – msd15213 May 5 '19 at 15:37
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    $\begingroup$ @mkmlp You only have to look at a few cases, the divisors of $24$. As for $n=2,4,8$, these groups are nilpotent, hence not simple. For the injection to $S_n$ see the answer here. $\endgroup$ – Dietrich Burde May 5 '19 at 15:37
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For the record, here is an elementary way of doing the first part - you might have another way of doing order $6$. It avoids any analysis of what the subgroups of order $8$ and $12$ actually are, but shows that if they exist, they have proper normal subgroups and therefore aren't simple.

The order of a subgroup is a factor of the order of the group.

Possible subgroup orders for $S_4$ are $12, 8, 6, 4,3,2$

The only groups of order $4,3,2$ are abelian.

A subgroup of order $6$ will contain an element of order $3$ and this will generate a normal subgroup (since the subgroup will have index $2$).

A subgroup of order $8$ will contain an odd permutation of order $2$ or $4$, and the even permutations will form a proper normal subgroup. (There aren't enough even permutations of order $2$ or $4$ to make a group of even permutations of order $8$)

A subgroup of order $12$ will either contain an odd permutation (in which case the even permutations form a proper normal subgroup) or will be all even permutations in which case the elements $e, (12)(34), (13)(24), (14)(23)$ will form a proper normal subgroup (union of conjugacy classes closed under multiplication and inverse).

The second part is covered by a link you have already been given.

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    $\begingroup$ You have missed out $24$ and $1$ from your list of possible orders of subgroups. $1$ is trivial, but you need to deal with $24$. Of course that just means showing that $S_4$ itself is not simple. $\endgroup$ – Derek Holt May 5 '19 at 16:14
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    $\begingroup$ @DerekHolt I was aware I hadn't dealt with those cases - you are right, of course, I should at least have done $24$ for which the even permutations form a normal subgroup. $\endgroup$ – Mark Bennet May 5 '19 at 17:01

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