Prove that there is no $\{3,5\}$-Hall subgroup in $A_{5}$ My approach.
A $\{3, 5\}$-Hall subgroup $K$ of $A_{5}$ has order $3\cdot 5$ and index $2^{2} = 4$. Note that $A_{5}$ acts in cosets of $K$, with $4$ distincts cosets, this way:
$$A_{5}/K = \{a_{1}K, a_{2}K, a_{3}K, a_{4}K\}.$$
Thus, we have a homomorphism $\varphi: A_{5} \to S_{4}$. Since $A_{5}$ is a simple group, $\ker \varphi = \{e\}$ or $\ker \varphi = A_{5}$. Then
Case 1: $\ker \varphi = \{e\}$.
So, $\ker \varphi = \{e\}$ implies $\varphi$ injective, an absurd because $|A_{5}| = 60 > 24 = |S_{4}|$.
Case 2: $\ker \varphi = A_{5}$.
So, for any $b \in A_{5}$ we have $b(a_{i}K) = a_{i}K$ iff $b \in a_{i}K$, where $1 \leq i \leq 4$, an absurd because $\bigcap(a_{i}K) = \emptyset$.
Therefore, there is no $\{3, 5\}$-Hall subgroup of $A_{5}$.

Is there an error? Or is it possible to improve this proof without using overkill results?

Even easier is to note that all groups of order 15 are cyclic, and $A_5$ has no element of order 15.  Thus $A_5$ has no subgroup of order 15.
