# $A_{4}$ unique subgroup of $S_4$ of order 12

I was reading the proof that $A_{4}$ is the unique subgroup of order $12$.

So the author counts the number of conjugates:

1 cycle of type () , 6 cycles of type $(1 \ 2)$, 8 cycles of type $(1 \ 2 \ 3)$, 6 cycles of type $(1 \ 2 \ 3 \ 4)$ and 3 cycles of type $(1 \ 2)(3 \ 4)$.

Now it says, the only possible way to get 12 elements is 1 + 3 + 8. Why is this? can't we have 1 cycle of type (), 2 cycles of type $(1 \ 2)$, 3 cycles of type $(1 \ 2 \ 3)$, 2 cycles of type $(1 \ 2)(3 \ 4)$ and 4 cycles of type $(1 \ 2 \ 3 \ 4)$. This also gives you 12. Why is this impossible though? I don't really understand why the only possibility is 1 + 3 + 8.

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Unique subgroup of what group? $S_4$? –  Myself Feb 17 '11 at 21:26
You cannot have $3$ cycles of type $(1,2,3)$, because each such cycle gives you two elements (the cycle and its inverse); they come in pairs. So the number of cycles of type $(1,2,3)$ in a subgroup must be even. –  Arturo Magidin Feb 17 '11 at 21:34

A subgroup of index 2 is always normal. Two elements of $S_n$ are conjugate iff they have the same cycle type. Thus, if your subgroup contains one element of a given cycle type, it contains all of them. Since the subgroup must contain the identity, and there's only one way of getting 11 as a sum from ${3,6,6,8}$, $A_4$ is the only possibility.