Proving if $|G|=280$, then $G$ is not simple

Suppose the group isn't simple $\;\implies n_5=56\implies\;$ there are $\;56\cdot 4=224\;$ elements of order $\;5\;$

Also, we have that $\;n_7\ge8\implies\;$ there are at least $\;8\cdot 6=48\;$ elements of order $\;7\;$ .

Thus, we already have $\;224+48=272\;$ elements of order $\;5\,\,\,or\;\;7\;$ . The ones left must belong to the then unique Sylow $\;2$-subgroup ...

Other proof's highlights . Suppose $\;n_5=56\;$ , then there exists a subgroup of $\;G\;$ with this order and this with index $\;\frac{280}{56}=5\;$ (namely, any Sylow $\;5$-subgroup's normalizer).

But then the action of $\;G\;$ on this subgroup's left cosets (i.e., the regular left action) renders a homomorphism $\;G\to S_5\;$ which, if $\;G\;$ is simple, must be injective. But this is absurd as $\;|G|\nmid 5!=120\;$ .