Show that $\theta$ is not the identity, yet $\theta^2$ is the identity (i.e. $\theta$ has order 2) where $\theta(a) = a^p$ (Sorry about the wording of the question, but that's the way it was phrased in the problem)
$p$ is a prime and $k$ is a field with $p^2$ elements. Define $\theta: k \to k$ by $\theta(a) = a^p$ for all $a \in k$.
Previous problems showed that $\theta$ is the identity if and only if $a = a^p$ for all $a \in k$, and that $\theta^2$ is the identity if and only if $a = a^{p^2}$ for all $a \in k$. The problem didn't state whether or not $\theta^2$ is $\theta(a)^2$ or $\theta(\theta(a))$, but I'm assuming it refers to the latter.
I think I have to show that if $\theta$ is not the identity, then $\theta^2$ must be the identity, which would mean that if $a \ne a^p$ then $a = a^{p^2}$. I'm not sure if it is possible to prove this, or if I'm supposed to go about this problem another way (i.e. showing that $\theta$ contains only 2 elements).
 A: Since we are working in a field (of characteristic $p$), it follows that the polynomial $x^p - x$ has (at most) $p$ roots. Since we have a prime field isomorphic to $\Bbb F_p$, and these are all roots of this polynomial, it follows that these are all the roots.
Therefore, for every $a \in k\setminus \Bbb F_p$, we have $a^p \neq a$. This shows $\theta$ is not the identity map.
On the other hand, we have that the non-zero elements of $k$ form a group of order $p^2 - 1$, from which it follows that for every non-zero element $b$, we have:
$\theta^2(b) = \theta(\theta(b)) = \theta(b^p) = (b^p)^p = b^{p^2} = b^{p^2-1}b = 1b = b$.
Since we also have $\theta^2(0) = 0$, it follows that $\theta^2 = \text{id}_k$.
As an example, let's look at $p = 2$, so that $k = \{0,1,a,a+1\}$, where $a$ is a root of $x^2 + x + 1$.
It is clear that $\theta$ is the identity on $\{0,1\}$, but for $a$:
$\theta(a) = a^2 = a+1$, so $\theta$ is not the identity.
On the other hand, $\theta^2(a) = a^4 = (a^2)^2 = (a+1)^2 = a^2 + 1 = (a + 1) + 1 = a$, and:
$\theta^2(a+1) = [(a+1)^2]^2 = (a^2 + 1)^2 = a^4 + 1 = a + 1$.
A: Since $k$ is a finite field of $p^2$ elements it is a splitting field of the polynomial $x^{p^2}-x$ over $\mathbf{Z}_p.$ Prove it if you did not know this fact already. 
Now for any $a\in\ k, a^{p^2}=\theta^2(a),$ whence $\theta^2(a)=a$ for any $a\in k.$ 
But can $\theta$ be the identity map? 
Consider the polynomial $x^p-x$. All of its roots are distinct as you can check that this polynomial and its derivative are relatively prime (you might need the fact that $\operatorname{char} k=p$). Can you see that $\theta$ fixes $\mathbf{Z}_p$ element-wise? That means each $a\in \mathbf{Z}_p$ is a root of $x^p-x.$ And $\mathbf{Z}_p$ has $p$ distinct elements. So any element in $k$ which is not in $\mathbf{Z}_p$ cannot be a root of $x^p-x.$ Hence $\theta$ cannot be the identity.
In fact $k$ is a finite field of $p^2$ elements if and only if it is a splitting field of the polynomial $x^{p^2}-x$ over $\mathbf{Z}_p.$
