Show that any group of order $2p$ is soluble Let $|G| = 2p$. The result is clear is $p$ is even.
The proof goes on to show that there is only one subgroup of order $p$ is p is an odd prime - my question is that, why do we need to show that there is only ONE subgroup of order $p$? By Sylow theorem, there exists a subgroup of order $p$, call this $H$ then $|H| = p$ so $H = C_p$ which is cyclic and hence abelian, and $|G / H| = 2$ so $G/H = C_2$ and is also cyclic and of index $2$ so is normal, so we have the chain $\{e\} \subset H \subset G$ where $H$ is normal in $G$ and $H$ and $G/H$ are cyclic, so abelian. 
Now in the proof I am given, they first use sylow theorems to show there is only one subgroup of order $p$, then they use my argument - my question is why is one subgroup needed, what would go wrong if there were multiple subgroups of order $p$?

The Sylow $p$-subgroup is unique iff it is a normal subgroup. But I agree that the argument that index 2 implies normal is more appealing.
