# 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?

Left multiplication by $$g$$ fixes the left coset $$hK$$ if and only if $$hK=ghK$$, i.e., $$g^h\in K$$ or equivalently $$g\in {^h}K$$. So in particular, since $$\ker\varphi$$ fixes $$K$$, it must be a subgroup of $$K$$ and hence cannot be $$A_5$$..
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.