<s>[INST] Find minimum polynomial of an element of $GF(2^5)$ I am trying to find the minimum polynomial of $\alpha^5$ and using this primitive polynomial to define my operations. $p(X)=1+X^2+X^5$.
The conjugates of an element $\beta$ of $GF(2^m)$ are $\{\beta^{(2^i)},\;i\geq 0 \}$.
The minimal polynomial is
$$\phi(X)=\prod_{i=0}^{e-1} (X+\beta^{2^i}).$$
$$\phi(X) = (X+\beta)(X+\beta^2)(X+\beta^4)(X+\beta^8)(X+\beta^{16})$$
I have set $\beta = \alpha^5$ and this gives me,
$$\phi(X)=X^5+X^4+(\alpha+\alpha^2+\alpha^4)X^3+(\alpha+\alpha^2+\alpha^3+\alpha^4)X^2+X+1$$
But the answer should be, $X^5+X^4+X^2+X+1$
Edit:
I have figured it out, It was a calculation mistake and I missed some terms. Now I have got the correct answer, i.e.
$$\phi(X)=X^5+X^4+X^2+X+1$$
 [/INST] Now that the OP found the mistakes in their calculation I will post my trick answer.
The element $\alpha^2$ is a conjugate of $\alpha$, so it shares with it the minimal polynomial $p(X)=X^5+X^2+1$. We can use this by observing that
$$\beta=\alpha^5=\alpha^5+p(\alpha)=1+\alpha^2.$$
Therefore $\beta$ must be a zero of
$$
q(X)=p(1+X)=(1+X)^5+(1+X)^2+1=X^5+X^4+X^2+X+1.
$$
After all,
$$
q(\beta)=q(1+\alpha^2)=p(1+(1+\alpha^2))=p(\alpha^2)=0.
$$
</s>