I'm a bit confused by part (a) and (d) specifically, in the following question,but I'll throw up the rest too just to make sure it's okay;
Let $β$ be a root $of g(x) = x^2 − x + 1$ over $GF(5)$.
(a) Explain briefly why $GF(5)(β) \cong GF(25)$.
(b) Verify that $β^3 \neq 1$ and $β^8 \neq 1$ in $GF(25)$.
(c) Using (b), or otherwise, find the minimum m ≥ 1 such that $β ^m = 1 $ in $GF(25)$.
(d) Determine the isomorphism type of the group (GF(25)×, ·) of non-zero elements of GF(25) under multiplication.
My answer:
a)$GF(5)$ is isomorphic to a subfield of $GF(25)=GF(5^2)$ as 1|2. $GF(5)(\beta):=\{a+b\beta|a,b \in GF(5)\}$ contains 25 elements so $GF(5)(\beta)\cong GF(25) $ as now it has the same number of elements as well as the same underlying structure. [I feel my explanation is shaky here].
b)$\beta^2-\beta+1=0 \Rightarrow\beta^2=\beta-1$.
so $\beta^3=\beta(\beta-1)=\beta^2-\beta=\beta -1-\beta=-1 \neq1$.
and $\beta^8=(\beta^2)^4=(\beta-1)^4=...=\beta^2$.
c)$\beta^{24}=(\beta^8)^3=(\beta-1)(\beta-1)(\beta-1)=(\beta^2-2\beta+1)(\beta-1)=(-\beta)(\beta-1)=-(\beta-1)+\beta=1$.
d) an automorphism ? [I really don't know about this one ]
For a), it depends on how you define $GF(5)$. My preferred argument is that $GF(5)$ is a splitting field: it's the splitting field for $t^{25}-t=0$ over $GF(5)$ and unique by uniqueness of splitting fields. But in any field of order 25, every element satisfies this polynomial, so it is also the splitting field.
For d), there's a standard result that any finite subgroup of the multiplicative group of a field is cyclic. The argument goes like this: Let $X$ be such a subgroup. It's certainly Abelian. Finite Abelian groups have elements of maximal order, and in fact every element has order dividing the maximal order: if $g$ has maximal order and $h$ has order not dividing $o(g)$, then $o(gh)=\frac{o(g)o(h)}{\text{gcd}(o(g),o(h))}>o(g)$, a contradiction.
So every element has order dividing $o(g)$, so every element is a root of the polynomial $x^{o(g)}-1=0$. There are certainly at least $o(g)$ elements in $\langle g \rangle$, so there are definitely that many in $X$. But there are at most $o(g)$ roots of this polynomial, so at most $o(g)$ elements of $X$. So $|X|=o(g)$, so $X=\langle g\rangle$.
In particular, this tells us that $GF(25)^{\times}$ is a cyclic group of order 24.