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.

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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.

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Let me make sure I am understanding this, So I see that is an automorphism on $G$, and since $H\subset G$, it must map every element in $H$ to an element that has order of a multiple of $5$, but since $|G|=15$, we know it must map to an element of order $5$, and every element of order $5$ is in $H$, so $ghg^{-1}\in H$ for all $h$. – tmpys May 10 '14 at 23:34
@tmpys It's simpler than that. We know that the image of a homomorphism is a subgroup, and that an automorphism is bijective. It follows that $\phi_g(H)$ is a subgroup of $G$ of order $5$. – Ayman Hourieh May 10 '14 at 23:36

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

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I don't know what Sylow theorems are. :( – tmpys May 10 '14 at 23:21
Here they are! – MattAllegro May 10 '14 at 23:23

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.

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I don't know what Sylow theorems are. :( – tmpys May 10 '14 at 23:22
That's quite unfortunate, because your question screams Sylow theorems. It'll be considerably harder to prove otherwise, but I'll give it some thought. – Kaj Hansen May 10 '14 at 23:26
I think this is leading us to Sylow theorems. – tmpys May 10 '14 at 23:36