# Every simple subgroup of $S_4$ is abelian

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.

• Please ask one question at a time. – Shaun May 5 '19 at 18:15
• @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. – msd15213 May 5 '19 at 18:17
• Fair enough :) ${}$ – Shaun May 5 '19 at 18:18
• @Shaun thanks ;) – msd15213 May 5 '19 at 18:19

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.

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

• I think the result you are using to answer the first question is more advanced than the question itself. – Derek Holt May 5 '19 at 15:16
• @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. – Dietrich Burde May 5 '19 at 15:17
• @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. – msd15213 May 5 '19 at 15:36
• @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. – msd15213 May 5 '19 at 15:37
• @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. – Dietrich Burde May 5 '19 at 15:37

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.

• 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. – Derek Holt May 5 '19 at 16:14
• @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. – Mark Bennet May 5 '19 at 17:01