How many sylow $3$-subgroups of $A_4$ are there? How many sylow $3$-subgroups of $A_4$ are there?
My attempt:
$n_3|O(A_4)\implies n_3=1,2,3,4,6,12$
$3|n_3-1\implies n_3=1,4$
How to calculate which of $1$ and $4$ is the value of $n_3?$
 A: You can also do this without using Sylow's theorems at all. Since $|A_4| = 12$, the Sylow 3-subgroups must have order 3, which means that each Sylow 3-subgroup is generated by an element of order 3.
Now, an element of order 3 must be a product of disjoint 3-cycles. Since we only have 4 points to act on, we can't squeeze more than one 3-cycle there. Therefore, an element of order 3 must simply be a 3-cycle.
Let's recap: we now know that each Sylow 3-subgroup is of the form $\langle a \rangle$, where $a$ is a 3-cycle. Now you can just look at all the $3$-cycles and the groups they generate. You'll see immediately that there are 8 cycles, and they generate 4 groups: $\langle (1\ 2\ 3) \rangle = \langle (3\ 2\ 1) \rangle$, $\langle (2\ 3\ 4) \rangle = \langle (4\ 3\ 2) \rangle$, $\langle (1\ 2\ 4) \rangle = \langle (4\ 2\ 1) \rangle$, $\langle (1\ 3\ 4) \rangle = \langle (4\ 3\ 1) \rangle$. So the answer is $4$.
A: Well just remember what $A_4$ is. Those are the permutations with signum 1. In particular every 3-cycle is in $A_4$. Furthermore for any $3$-cycle $\pi$ you know that 
$$\{id, \pi,\pi^{-1}\}$$ 
is a subgroup with cardinality $3$. 
In particular in $A_4$ there are $(1\ 2\ 3)$ and $(1\ 2 \ 4)$, so you have at least $2$ different $3$-Sylow subgroups.
A: Hint: If $n_3=1$, then the unique element $P$ of $\text{Syl}_3(A_4)$ would necessarily be normal. Let $x\in A_4$ be of order $2$. Then, by the normality of $P$, we know that $\langle x\rangle P\leqslant A_4$. But,
$$|\langle x\rangle P|=\frac{|\langle x\rangle||P|}{|P\cap\langle x\rangle|}=\frac{2\cdot 3}{1}=6$$
What's that famous property of $A_4$? :)
A: The $3$-subgroups of $A_4$ are determined by the one element of $\{1,2,3,4\}$ that is left fix by the subgroup under the group action. Hence there are $4$ such subgroups.
A: $A_4$ is easily shown to be the rotation symmetry group of the regular tetrahedron. About each vertex is a cyclic rotation symmetry.
