<s>[INST] Check: Find all the Sylow $3$-subgroups of $A_4$. Find all the Sylow $3$-subgroups of $A_4$.
The order of $A_4$ is $\frac{4!}{2}=\frac{24}{2}=12=3\cdot 2^2$. So the Sylow $3$-subgroups are


*

*$\{ \begin{pmatrix} 1 \end{pmatrix}, \begin{pmatrix} 2 & 3 & 4 \end{pmatrix}, \begin{pmatrix} 2 & 4 & 3 \end{pmatrix} \}$

*$\{ \begin{pmatrix} 1 \end{pmatrix}, \begin{pmatrix} 1 & 2 & 3
    \end{pmatrix}, \begin{pmatrix} 1 & 3 & 2 \end{pmatrix} \}$

*$\{ \begin{pmatrix} 1 \end{pmatrix}, \begin{pmatrix} 1 & 2 & 4
    \end{pmatrix}, \begin{pmatrix} 1 & 4 & 2 \end{pmatrix} \}$

*$\{ \begin{pmatrix} 1 \end{pmatrix}, \begin{pmatrix} 1 & 3 & 4 \end{pmatrix}, \begin{pmatrix} 1 & 4 & 3 \end{pmatrix} \}$


Should I add more details or is this sufficient?
 [/INST] The highest power of 3 that divides $|A_4|$ is $3^1=3$.  So we need to list all subgroups of $A_4$ of order 3.  One of them is clearly $\langle (234) \rangle$, as you mention.  Now we need to determine all the remaining subgroups of order 3.  One way is as follows. Since a group of order 3 is cyclic, a subgroup of order 3 is of the form $\langle (ijk) \rangle$.  The number of ways to choose 3 elements from 4 elements is ${4 \choose 3}=4$.  Once the three elements are chosen, the three-cycle and its inverse that (along with the identity) form the subgroup of order 3 are uniquely determined. 
We could also use the part of Sylow's theorem that says that all Sylow 3-subgroups are conjugate. Hence they all can be obtained by relabeling the points in the first subgroup $\langle (234) \rangle$. This is because conjugation effects a relabeling of the points.  
Alternatively, once you list (using any method) the four subgroups you mention, we know these are all the Sylow 3-subgroups. For if a group $G$ has order $p^k m$, where $p$ does not divide $m$, then the number $n_p$ of Sylow $p$-subgroups is congruent to 1 mod $p$ and $n_p$ divides $m$. In this special case, the number $n_3$ of Sylow 3-subgroups is congruent to 1 mod 3 and must divide 4, so $n_3$ must equal 1 or 4. But since you found more than one Sylow 3-subgroup, $n_3$ equals 4.
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