Nonabelian group of order $p^3$ for odd prime $p$ and exponent $p$ I am trying to prove the following:
Let $P$ be a nonabelian group of order $p^3$ where $p$ is an odd prime and assume that $P$ has exponent $p$. Then $\text{Out}(P) \cong GL_2(p)$.
My teacher have tried to give me an idea of how we can look at $P$ - as a Sylow $p$-subgroup of $(C_p\times C_p) \rtimes GL_2(p)$. The reasoning is as follows (and I don't fully understand every step, but you will see that):
Since $P$ has exponent $p$, we know that every nontrivial element of $P$ has order $p$. We know that $Z(P)\neq 1$ since $P$ is a $p$-group, and since $P$ is nonabelian we know that $Z(P)\neq P$. If we assume that $\vert Z(P)\vert = p^2$ then $G/Z(G)$ is cyclic making $G$ abelian, so we conclude that $\vert Z(P)\vert = p$.
Now let $x\in P\backslash Z(P)$ and set $U = Z(P) \langle x\rangle$ Then $U \cong C_p \times C_p$. 
First question: How do I know that $Z(P) \langle x\rangle \cong C_p \times C_p$. I mean, why is $Z(P)\cap \langle x\rangle = 1$ and how do I know that $Z(P)\langle x\rangle$ is not cyclic?
Ok, well $U$ has index $p$ in $P$ and is thus a normal and maximal subgroup of $P$. Let $y\in P\backslash U$. Then $P = U \rtimes \langle y\rangle$.
Second question: How do I know this? I mean, I know that $U$ is a normal and maximal subgroup of $P$ and $y\notin U$, so $U \langle y\rangle$ is a subgroup of $P$ containing $U$ properly, ergo $U \langle y\rangle = P$. But why is $U \langle y\rangle = U \rtimes \langle y\rangle$?
Proceeding, I realize that $\text{Aut}(U) \cong GL_2(p)$, so $\vert \text{Aut}(U) \vert = p(p-1)(p^2-1)$, so a Sylow $p$-subgroup of $U$ has order $p$. And therefore, we can see $P$ as a Sylow $p$-subgroup of $U\rtimes GL_2(p)$.
Third question: How do I get that $P$ is a Sylow $p$-subgroup of $U\rtimes GL_2(p)$ from the reasoning above?
And
Final question: How can I use this knowledge to prove that $\text{Out}(P) \cong GL_2(p)$?
I know this is a lot, but just a little help is appreciated!
 A: The fact is that there's just one such a group. And it is
$\{A \in GL_3(\mathbb{F}_p)| a_{i,j}=0, i<j, a_{i,i}=1\}$. This group works because of the Frobenius and because p is at least 3(as the maximum order of nilpotence of those matrices minus Id).
Now consider the following abstract group given by generator and relations:
$G=\{x,y,z|x^p=y^p=z^p=1,xy=yx,zx=xz,yz=zyx\}$.
Using the relations you find that this group has order at most $p^3$, since the relations immediately implies that you can write any element as $x^ay^bz^c,0 \leq a,b,c \leq p-1$.
Now take any non abelian group H of order $p^3$ with exponent p. Take two non central element y',z', since $G/H \simeq (\mathbb{Z}/p\mathbb{Z})^2$,you can choose them as independent vectors in the quotient,so you have a relation $y'z'=z'y'x'$ with x' a central element not 1(otherwise H is abelian), and so a generator of the center. So consider the map $G \to H$ that sends $x \to x',y \to y',z \to z'$. Since the relation are preserved this extends uniquely to a surjective group homomorphism. So since $|G| \leq p^3$ but the map is surjective, we conclude that if there exists such H, then $|G|=p^3$ and moreover any group of order $p^3$ non abelian and of exponent p is isomorphic to G.
But since we have exibithed such a group we have that it is the only one up to isomorphism and we have also a presentation by generator and relations.
Now given that your group has the presentation of G, the thesis follows immediately:
$G/Z(G) \simeq (\mathbb{Z}/p\mathbb{Z})^2$. So any automorphism gives an automorphism of the quotient, so you get a map $Aut(G) \to GL_2(\mathbb{F}_p)$. Now this map factors trough inner automorphisms, and gives you your isomomorphism. Indeed, the image has $(p^2-1)(p^2-p)$. It's easy to show that the map is surjective. And you can also prove that the number of automorphism is at most $p^2(p^2-1)(p^2-p)$. So the kernel has to be exactly the inner automorphisms. And you get an isomorphism between $Out(G)$ and $GL_2(\mathbb{F}_p)$.
A: First question. The group generated by the center and $x$ non-central is obviously (...) commutative of order $>p$ and $<p^3$ since $P$ is noncommutative, so it is of order $p^2$. There are, up to isomorhism, only two groups of order $p^2$, the cyclic group and $\Bbb Z/p\oplus\Bbb Z/p$. The first one is ruled out by the hypothesis that the exponent of $P$ be $p$.
Second question. $P$ is the semidirect product of $U\simeq\Bbb Z/p\oplus\Bbb Z/p$ and $\langle y\rangle\simeq\Bbb Z/p$ since you have a split exact sequence
$$1\to U\to P\stackrel{\curvearrowleft}{\to}P/U\to 1$$
where the splitting is given by $\sigma:U/P\simeq\Bbb Z/p\to P,1\mapsto y$. The fact that$\sigma(P/U)$ meets $U$ trivially comes from the fact that their intersection is a subroup of $\langle y\rangle$ so it is either $1$ or all of $\langle y\rangle$, which would mean that $y\in U$ contrary to the assumption on $y$, so the intersection is trivial and $\sigma$ is a splitting.
Third question. Suppose you have groups $h,H$ and $N$, where $h\subset H$ is a subgroup of $H$, and a morphism of groups $H\to\mathrm{Aut}(N)$. Then the canonical map $$N\rtimes h\to N\rtimes H$$ is an embedding of groups. Here you have an embedding $\langle y\rangle=h\hookrightarrow H=\mathrm{Aut}(U)\simeq\mathrm{GL}_2(p)$ where $y$ acts on $U$ by conjugation ($y\cdot u=yuy^{-1}$, it is indeed an embedding, otherwise $y$ would commute to $x$ and $P$ would be commutative). Thus, the statement above yields an embedding
$$P=U\rtimes\langle y\rangle\hookrightarrow U\rtimes\mathrm{Aut}(U)\simeq U\rtimes\mathrm{GL}_2(p)$$
The fact that this embeds $P$ as a Sylow $p$-group of the right-hand side follows from cardinalities.
Final question. I might get back to this later.
A: I think there are some elementary computation here. As Geoff Robinson explain, the Sylow subgroup of $C_p \times C_p \rtimes GL_2(p)$. Since the sylow subgroup of $GL_2(p)$ can be generated by 
$\alpha=\left ( \begin{array}{cc}
1 & 1\\
0 &1
\end{array}
\right ),
$
we can get the presentation of $P$ as $\langle z, u, \alpha \rangle$, where $z^p=u^p=\alpha^p=1$, $[z,u]=1, [z,\alpha]=1$,and $[u,\alpha]=z$. 
Note $f \in Aut(P)$ $f(z), f(u), f(\alpha)$ satisfy the relation of $z,u,\alpha$. In fact only $[u,\alpha]=z$ need to pay attention in this case. Since the nilpotency class of $P$ is 2, this can be computed easily.
Now $A=Aut(P)$ acts on $P/\langle z \rangle$. Let $g$ is in the kernel of this action. We get $f: u\rightarrow z^s u$ and $ \alpha \rightarrow z^t \alpha$. Suppose $f:z\rightarrow z^r$. From $[f(u), f(\alpha)]=f(z)$, we get $r=1$ and $s,t \in F_p$. This means the kernel $K$ is of order $p^2$. Clearly, the element in $Inn(P)$ acts on $P/\langle z \rangle$ as the $f \in K$. We get $K=Inn(P)$ for $Inn(P)|=p^2$. So we get, as jimmy page said, $Out(P)=A/Inn(P)=A/K$ become a subgroup of $GL_2(p)$.
Now $f: u\rightarrow u^a\alpha^c, \alpha \rightarrow u^b\alpha^d$ for $a,b, c,d \in F_p$ and $ad-bc=r \ne 0$, then we can let $f:z\rightarrow z^r$, and find $[f(u), f(\alpha)]=f(z)$. Hence such $f \in Aut(P)$. This means $|Out(P)|\ge |GL_2(p)|$.
Now we get $Out(P) \cong GL_2(p)$. 
