Problem as in the title, where $p$ is prime. I know that it has to be cyclic, and I have proved that it’s $C_1$ iff $p\equiv 1 \pmod 5$. The full answer that, and $C_2$ if $p\equiv 4 \pmod 5$, and $C_4$ if $p\equiv 2,3\pmod 5$.
How can I prove the latter two cases? I also know that this is equivalent to finding when $x^4+x^3+x^2+x+1$ is irreducible.
On
Recall that if $f(x) \in \mathbb{F}_p[x]$ is any polynomial, the Frobenius map $F : x \mapsto x^p$ generates the Galois group of its splitting field, and hence to compute its Galois group it suffices to compute the cycle structure of Frobenius acting on its roots; the Galois group will be cyclic of order the lcm of the sizes of the cycles.
When $f(x) = x^n - 1$ and $\gcd(p, n) = 1$ the roots are precisely powers of a primitive $n^{th}$ root of unity $\zeta_n \in \overline{\mathbb{F}_p}$ and so we can be very explicit about the action of Frobenius on it: we have $F(\zeta_n) = \zeta_n^p$ and so $F^k(\zeta_n) = \zeta_n^{p^k}$, meaning that the orbit of $\zeta_n$ has size the least positive integer $k$ such that $\zeta_n^{p^k} = \zeta$, or equivalently such that
$$p^k \equiv 1 \bmod n.$$
This is exactly the multiplicative order $\text{ord}_n(p)$ of $p \bmod n$. The other roots of $f(x)$ are the other $n^{th}$ roots of unity $\zeta_n^k$, which are primitive $\frac{n}{\gcd(n, k)}$ roots of unity, and hence which have orbits of size $\text{ord}_{\frac{n}{\gcd(n, k)}}(p)$, which in particular divides the size of this largest orbit we found above.
Hence the Galois group has order $\text{ord}_n(p)$, but the analysis above even reveals the exact cycle structure of Frobenius, and furthermore reveals it on each of the irreducible factors
$$x^n - 1 = \prod_{d | n} \Phi_d(x)$$
of $x^n - 1$ over $\mathbb{Q}$ (the cyclotomic polynomials).
On
Your problem can be nicely generalized as follows: Suppose that p, q two distinct primes such that the class p mod q is a generator of the multiplicative group $\mathbf F_q^{\times}$ ; in other words, for any $d < q-1,q$ does not divide $p^d - 1$. Then $f_q(X):= X^q - 1/X - 1$ is irreducible over $\mathbf F_p$. In your case, $q=5$ and it is immediately checked that the hypothesis is equivalent to $p \neq \pm 1$ mod $5$.
Proof: It is a simple application of the Galois theory of finite fields. Let $\zeta_q$ be a primitive $q$-th root of unity (in an algebraic closure of $\mathbf F_p$). The extension $E=\mathbf F_p(\zeta_q) =$ splitting field of $f_q(X)$ over $\mathbf F_p$, has the form $\mathbf F_{p^d}$, where $d$ is the degree of the minimal polynomial of $\zeta_q$ over $\mathbf F_p$, in particular $d \le (q-1)$. Besides, $E^{\times}$ is cyclic of order ($p^d-1$), hence $q$ divides $(p^d - 1)$; by our hypothesis, it follows that $d \ge q-1$, so finally $d=q-1$ and $f_q(X)$ is irreducible.
The polynomial $f=x^4+x^3+x^2+x+1$ is irreducible over $\Bbb F_p$ if $p\not\equiv \pm 1\bmod 5$ with $p\neq 5$. The proof goes as follows. Note that $$ x^5 −1 = (x−1)(x^4 +x^3 +x^2 +x+1). $$ Any root of $f=0$ has order $5$ or $1$ (in a splitting field). Only the identity has order $1$. If $f$ had a linear factor in $\Bbb F_p[x]$, then $f=0$ had a root in $\Bbb F_p$. Since $1+1+1+1+1\neq 0$, $1$ is not a root, so that any possible root must have order $5$. But the order of $\Bbb F_p^*$ is $p-1$, which is not divisible by $5$, so there is no root in the base field $\Bbb F_p$.
If $f$ had an irreducible quadratic factor $q$ in $\Bbb F_p[x]$, then $f=0$ would have a root in a quadratic extension $K$ of $\Bbb F_p$. Since $[K : \Bbb F_p] = 2$, the field $K$ has $p^2$ elements, and $K^*$ has $p^2 −1=(p−1)(p+1)$ elements. By Lagrange, the order of any element of $K^*$ is a divisor of $p^2 − 1$, but $5$ does not divide $p^2 − 1$, so there is no element in $K$ of order 5. That is, there is no quadratic irreducible factor.
It follows that $f$ is irreducible.