How do i prove that $A_4$ has a unique Sylow 2-subgroup Let $A_4$ be the alternating group pf degree 4.
How do i prove that $A_4$ has a unique Sylow 2-subgroup?
 A: There are very few elements (relatively) in $A_4$, to wit, $4!/2=24/2=12$ elements. Verify there are exactly $4$ elements of order $2$, that constitute the only Sylow subgroup of $A_4$, and that it is isomorphic to $V_4$ the Klein four group. These elements are the double transpositions $(12)(34),(14)(23),(13)(24)$ and $1$. Note that since any pair of $2$-Sylows are isomorphic, any other $2$-Sylow must consist of elements of order $2$. Since the ones mentioned are all of them, there are no other $2$-Sylow subgroups.
A: We know that sylow $2$ subgroup of $A_4$ has $4$ elements and  $A_4$ does not include any elemets of order $4$. 
We need to count the elements of the form $(a,b)(c,d)$,
${\dfrac{1}{2}}{4\choose2}.{2\choose2}=3$ we divide by $2$ as $(a,b)(c,d)=(c,d)(a,b)$.
And with identity we have $4$ elements so we have uniqe sylow $2$ groups.
A: Another way is to count the number of Sylow $3$-subgroups. There must be more than one, since $(234)$ is not a power of $(123).$ Hence there must be at lest $4$ Sylow $3$-subgroups. This gives at least $8$ elements of order $3$. Hence there are at most $4$ elements whose order is a power of $2.$ Since a Sylow $2$-subgroup has order $4,$ these elements form a subgroup of order $4,$ which must be the unique Sylow $2$-subgroup.
A: Let $n_2$ be the number of Sylow $2$-subgroups of $A_4$. The Sylow theorems state that $n_2\equiv1\pmod2$. Since $|A_4|=4!/2=12=2^2\cdot 3$, it follows that $n_2=1$ or $n_2=3$.
Now, seeking a contradiction, suppose that $n_2=3$ and let $G$ be a Sylow $2$-subgroup. The Sylow theorems also state that $n_2=|A_4|/|N_{A_4}(G)|$ where $N_{A_4}(G)$ is the normalizer of $G$ in $A_4$. It follows that $3=12/|N_{A_4}(G)|$ so that $|N_{A_4}(G)|=4$. But $|G|=2^2=4$ so $G=N_{A_4}(G)$. That is, $G$ is normal in $A_4$, a contradiction since $G$ is conjugate to the other two Sylow $2$-subgroups of $A_4$.
Hence $n_2=1$ as desired.
