I have been given the problem of determing $n_2(A_5)$.
I do not want the answer to this, since I am making some progress by myself. The issue is, I have run into what seems like a contradiction, and I can't figure out where I went wrong.
We start by observing that $|A_5|=60=2^2\cdot 3\cdot 5$. This means that any Sylow 2-subgroup is going to be of order 4. Furthermore, properties of $A_5$ let us conclude that there are no elements $\tau \in A_5$ with $|\tau|=4$.
Therefore, we conclude that none of the Sylow 2-subgroups of $A_5$ are cyclic. Thus, each one is isomorphic to the Klein 4-group. Since the only elements of $A_5$ are pairs of disjoint 2-cycles, I wrote out the 2-subgroups as follows:
$$ P = \{1, (a\ b)(c\ d),(a\ b)(e\ f),(c\ d)(e\ f)\} $$ for distinct $a,\dots,f$, since this will satisfy the properties for the Klein 4-group. However, we are working in $A_5$, so we can't actually pick distinct $a,\dots,f$ to construct this.
Where did I go wrong here?
Permutation groups are highly non-commutative. If you choose any two permutations in random, then they do not commute. In particular, if you are picking two permutations of the type $(a,b)(c,d)$, then they may not commute, hence do not fall in the (abelian) $K_4$ group. What should be done?
Note that $A_5$ has 5 copies of $A_4$: each copy is obtained by considering all the even permutations which fix a given letter.
Then, in each $A_4$, you have Sylow-$2$ subgroup - it is $K_4$.
This give all the Sylow-$2$ subgroups of $A_5$ (can be proved by some group-action argument or other way one thinks.)