I know that A4 is a subgroup of order 12. So I tried to prove that S4 has a unique subgroup of order 12 by assuming that there existed another subgroup of order 12, lets call this other subgroup H. The hint I was given was to also assume that the subgroup H was normal. Any tips on where I should go from here?
Now, permutations in are special. A permutation of a specific cycle type* is conjugate to every other permutation of that cycle type. (g is conjugate to h if there exists some element a such that ). For example, . This should hopefully be in your notes. If it is not, then say and someone will have to come up with a different method to solve this probelm!
A normal subgroup is a group which is closed under conjugation. So in , if there exists a permutation of a cycle type then all permutations of this cycle type are in the normal subgroup.
Let us now count the number of each cycle types,
There is 1 cycle of type in .
There are 4.3/2 = 6 cycles of type in .
There are 4.3.2/3 = 8 cycles of type in .
There are 126.96.36.199/4 = 6 cycles of type in .
There are 188.8.131.52/(2.2.2) = 3 cycles of type in .
(for a normal cycle you divide by the length as , etc. For the last part you divide by 2.2.2 as two of the twos are because and the third is because . Alternatively, just sum up all the other cycle types and subtract from 24, 24-(6+8+6+1) = 3.)
Note that these all sum up to 24. Now, we wish to sum together the numbers 1, 6, 8, 6 and 3 to get 12 in two different way. However, there is only one way of doing this. This means there can only be one subgroup of order 12 in .
*Note that by cycle type I mean , , are all different while and have the same cycle type.