# How can I prove that there exists only one subgroup of A4 of order 4

Size of A4 =12

So, there are either 1 or 3 2-sylow subgroups

and there are either 1 or 4 3-sylow subgroups.

Suppose that there are 4 3-sylow subgroups. Then, there are 8 elements of order 3

Thus, 2-sylow subgroup must be unique.

However, this is case when I suppose that there are 3-sylow subgroups.

What If I suppose that there are 2-sylow subgroups and derive the result that there can't be 4 3-sylow subgroups?

Also, How can I find the subgroup of order 4?

Do I have to look at every element in A4 and consider possible cases?

Looking at the elements of $A_4$ is not that difficult. If you know what the elements of $S_4$ look like (identity, $2$-, $3$-, $4$-cycles and products of two disjoint transpositions (i.e. $2$-cycles)), pick the ones with even parity: $(1)$, $(a \ b \ c)$ and $(a \ b)(c\ d)$, which have orders $1$, $3$ and $2$ respectively. Since there are no elements of order $4$, each Sylow $2$-subgroup consists of the identity and $3$ elements of order $2$. It's not difficult to see that there are only $3$ elements of the form $(a \ b)(c\ d)$. Hence you can only form one subgroup of order $4$.