Show that any group of order $2p$ is soluble

Question

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 $p$?

Answer 1

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.

Answer 2

You don't need the full strength of Sylow I to show that there is a subgroup of order $ p $, using Cauchy's theorem suffices. As an alternative way of proving uniqueness, let $ H $ be our cyclic subgroup of order $ p $ and let $ g $ be an element of order $ p $. Now, let $ \langle g \rangle $ act on $ G/H $ (the left coset space) by left multiplication. By fixed point congruence, $ \textrm{Fix}_{\langle g \rangle}(G/H) \equiv |G/H| = 2 \pmod{p} $, so that the entire set $ G/H $ is fixed by the action, and in particular we have $ gH = H $ or $ g \in H $.

On the other hand, as you've remarked, this is not necessary. It does, however, show that this is the only possible composition series of $ G $, which perhaps is the point of proving that the subgroup is unique.