How to show $A_4$ is the only subgroup of order 12 of $S_4$ ? (Do not use Lagrange, index, cosets, normal subgroups, quotient groups)

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Context: Actually, this post is a specific version of my other question: If I happen to have like a subgroup of $S_4$ that is generated by 2 (distinct) elements $a,b$ and then after much multiplication I arrive at 10 more elements, for a total of 12 elements, then how do I go about proving that there aren't any more elements?

There's this question $A_n$ is the only subgroup of $S_n$ of index $2$. but it appears each answer involves one of the following things I'm not allowed: Lagrange, index, cosets, normal subgroups, quotient groups.

This has been asked before specifically for $n=4$, but it appears each question or answer involves one of the above things.

  1. $S_4$ has only a subgroups of order 12?

  2. Prove $S_4$ has only 1 subgroup of order 12

  3. $A_{4}$ unique subgroup of $S_4$ of order $12$

Guess: The only thing that appears to be left is to do the multiplication table, i.e. multiply each element by itself and then the other 11 to get that each of the 144 products are all one of the 12 elements.

Best Answer

Hints.

Let $H$ be a subgroup in $S_4$ of order $12$ different from $A_4$. Then $H$ contains an odd permutation. It follows that the number of even and odd permutations in $H$ is equal (I'm sure you can easily prove this). Hence, the even permutations of $H$ form a subgroup $F$ of order $6$ (why). Further we consider two cases:

  1. $F$ contains two cycles of length $3$ of the form $(123)$ and $(124)$;
  2. $F$ does not contain a pair of such cycles, then $F$ contains all three permutations of the form $(12)(34)$.

In each of these cases it is necessary to multiply the permutations a little. But this is not much work.

Edit.

For example, in the first case. In the subgroup $F$ must lie permutations: $e,(123),(132),(124),(142),(123)(124)=(14)(23),(123)(142)=(234)$. There is no need to compute further. We already have $|F|>6$. Contradiction.

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