Find the number of $2$-Sylow of $D_{10}$
I got this problem of group theory. I tried to solve it this way:
Using third Sylow theorem, we know that the number of $5$-Sylow is $1$, so we have only one subgroup $H$ of order $5$ of $G=D_{10}$, with $|G|=20$. Also, we have that the number of $2$-Sylow is either $1$ or $5$, so if $n_2=1$, we have only one subgroup $K$ of order $4$, and we know that both $H$ and $K$ are normal in $G$, so $G$ is isomorphic with the external direct product of its Sylow normal subgroups, so it's either $\Bbb Z_4\times\Bbb Z_5$ or $\Bbb Z_2\times \Bbb Z_2\times\Bbb Z_5$, but in both cases $G$ would be abelian, so a contradiction.
Is it right?
Thank you for all of your support.