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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?

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That is correct (and exactly how I would have proved it).

As for improvement, I think in Case 2 it would be sufficient to say that the action on cosets of a subgroup is always transitive, and thus the kernel of a coset action is always a proper subgroup.

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  • $\begingroup$ Really! Thanks for the hint! $\endgroup$
    – Corrêa
    Sep 29 '18 at 2:19
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Your case 2 doesn't look right.

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 $A_5$..

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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.

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