Question on proof $G$ is not simple

Question

I have a question on the proof of: Proving if $|G|=280$, then $G$ is not simple

$n_7=8$ so there are $48$ elements of order $7$.

$n_5=56$ so sthere are $224$ elements of order $5$.

Now, in the proof, the $8$ elements left belong to the $2$ Sylow group, so it follows that there is only one $2$ Sylow subgroup, which is normal, too. How does it follow?

Is it also right, if I argue with there are at most $35$ elements of order $2$, so there are $48+224+35+1=308>280$ elements, so $G$ can't be simple?

Answer 1

Every element which lies in some Sylow $2$-subgroup has order equal to a power of $2$. Since the Sylow $2$-subgroups of $G$ have $8$ elements each, if there was more than one, then there would be more than $8$ elements of order a power of $2$, contradicting what was already established.