Show that a group of order fifteen has an normal subgroup of order 5 Let $G$ be a group where $|G|=15$, I want to show that $G$ has a normal subgroup of order $5$.
I have shown that $G$ must have a subgroup $H$ of order $5$, (and one of order $3$), and I have shown that there is only one subgroup of order $5$, the book says the rest should be obvious, but it is not to me...any help would be great.
 A: For a fixed $g \in G$, consider $\phi_g(x) = gxg^{-1}$. This is an automorphism of $G$. It must map $H$ to a subgroup of $G$ of order $5$. But you've shown that there is only one such subgroup. Thus $gHg^{-1} = H$ for all $g \in G$. It follows that $H$ is normal.
A: If you can show that there exists only $1$ Sylow 5-subgroup, then that subgroup of order $5$ is necessarily normal since all Sylow $p$-subgroups are conjugate to one another for a given $p$ (by Sylow II).
It is actually possible to prove an even stronger statement.  Here's a rough sketch:
Using the Sylow theorems, any group of order $15$ must have exactly one Sylow $3$-subgroup and exactly one Sylow $5$-subgroup.  It is possible to show that any $G$ such that $|G|=15$ is isomorphic to a direct product of these two Sylow subgroups.  Since the Sylow subgroups are of prime order, we have $G \cong \mathbb{Z}_3 \times \mathbb{Z}_5 \cong \mathbb{Z}_{15}$.  Thus, any group of order $15$ is cyclic, and we conclude that all of its subgroups are normal.
A: By Sylow theorems, all Sylow p-subgroups are conjugate to each other: If H and K are Sylow p-subgroups of G, then there is some $ g \in G $ such that $ H = g^{-1}Kg $.
So, if you know there is only one Sylow 5-subgroup H, it must be that for all $ g \in G $, $ H = g^{-1}Hg $ and hence $ H $ is normal.
Edit: Wikipedia link to the statement of Sylow's theorems: http://en.wikipedia.org/wiki/Sylow_theorems#Theorems
