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Proof attempt: We complete this problem through application of Sylow's Theorems. Note that $|G| = 60 = 2^2\cdot 3\cdot 5$. We have the following possibilities for the number, $h_p$, of Sylow $p$-groups: \begin{align*} h_5 &= 1, \ 6 \\ h_3 &= 1, \ 4, \ 10 \\ h_2 &= 1, \ 3, \ 5, \ 15 \end{align*} Since $G$ is simple, we can immediately eliminate 1 as an option for all $h_p$. It follows that $h_5 = 6$. Next, we examine the possibilities for $h_3$. Either $h_3 = 4$ or $h_3 = 10$. Suppose that $h_3 = 4$. If this is the case, then there exists a homomorphism of $G$ into $S_4$. Since $|S_4| = 24$ and $|G| = 60$, the kernel of this homomorphism is non-trivial, meaning that there exists a non-trivial normal subgroup of $G$. This contradicts $G$ being simple, and it follows that $h_3 = 10$.

Lastly, we look at $h_2$. An identical argument to the one given for $h_3$ removes 3 as a possibility for $h_2$. This leaves us with either 5 or 15 as the only reasonable values for $h_2$. Currently, we know that there are $6\cdot 4 + 10\cdot 2 = 44$ elements in our group.

At this point, I want to show that assuming $h_2 =15$ leads to a contradiction (namely, that the group has more than 60 elements), but I'm not sure how to proceed. Because the Sylow $2$-subgroups have order 4, they may have non-trivial intersections. I want to show that it's somehow impossible for each 2-subgroup to contribute only one unique element of order 2. If I can do that, $h_2=5$ and the result will follow. Any help is appreciated!

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up vote 4 down vote accepted

Although $h_2 = 15$ is not actually possible, try to still show that in this case there is a subgroup of index $5$. It is not possible that in this case intersections of different Sylow $2$-subgroups are trivial, and thus there is a subgroup $K$ of order $2$ normalized by two Sylow $2$-subgroups. What can the order of $N_G(K)$ be then?

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