Let $R = \mathbb F_3[X]/\langle X^3 + X^2 + 1\rangle$ and $\alpha = [X] \in R$. Show $R^*$ is not cyclic. 
Let $R = \mathbb F_3[X]/\langle X^3 + X^2 + 1\rangle$ and $\alpha = [X] \in R$. Show $R^*$ is not cyclic.

I've proven:
$X^3+ X^2 + a \in \mathbb F_3[X]$ is irreducible $\iff$ $a=2$.
$\alpha \in R^*$ and ord($\alpha$)=$8$ with respect to the group $(R^*, \cdot)$.
$\alpha^4$ and $-\alpha^4$ are two different elements in $R^*$ and ord($\alpha^4$)=ord($-\alpha^4$)=$2$ with respect to the group $(R^*, \cdot)$.
To prove:
$R^*$ is not cyclic.
Ideas:
I know $|R|=3^3 = 81$ and $R$ is not a field. The numbers $2$ and $8$ divides |R^*|, $|R^*| > 8$ which implies $|R^*| = 8k$ for $2 \le k \le 10$. $\mathbb F_3 \cong \{0,1,2\} \in R$.
I find these cyclic exercises very difficult, are there some general guidelines in solving them ?
 A: 
Let $G$ be a cyclic group, then every subgroup of $H \le G$ is also cyclic by a theorem that should sound familiar to most people.

Let $H = \{\alpha^4, -\alpha^4, 1,-1\} \subseteq R^*$. Then by verification $H$ is indeed a subgroup of $(R^*, \cdot)$, since $1 \in H, (\alpha^4)^{-1} \in H, (-\alpha^4)^{-1} \in H$ and for any two elements in $H$ the product also lies in $H$ (ie. $\pm \alpha^8 = \pm 1$ .
Now in the question above it has been proved that the order of $\alpha^4$ and $-\alpha^4$ is $2$. By looking at $-1$ we see it has order $2$ also. The order of $1$ is $1$.
Thus no element in $H$ has order 4, which implies by contraposition, $R^*$ cannot be cyclic, since it has a subgroup which is not cyclic.
A: You have already seen that $X^3 + X^2 + 1$ is reducible: $X^3 + X^2 + 1 = (X + 2)(X^2 + 2X + 2)$ with $X^2 + 2X + 2$ being irreducible. This allows you to complete determine $R$ and also $R^*$.
By the Chinese Remainder Theorem, $R \cong {\mathbb F}_3[X]/(X + 2) \times {\mathbb F}_3[X]/(X^2 + 2X + 2)$, so $R \cong {\mathbb F}_3 \times {\mathbb F}_9$. Therefore $R^* \cong {\mathbb F}_3^* \times {\mathbb F}_9^* \cong C_2 \times C_8$. This last group it is not cyclic: it has order 16, but the order of every element is at most 8.
Edit. For an alternative reasoning that $R^*$ is not cyclic, without using the total structure of $R$, note that a cylic group has either 0 or 1 elements of order 2. However, in $R^*$, $2$, $1 + 2\alpha + \alpha^2$, and $2 + \alpha + 2\alpha^2$ all have order 2.
(Of course, this is easier to find once you know that $R \cong {\mathbb F}_3 \times {\mathbb F}_9$, because in ${\mathbb F}_3 \times {\mathbb F}_9$ these are the elements $(-1,-1)$, $(1,-1)$ and $(-1,1)$.
This is also essentially the same argument as the OP gives himself in his answer: $\alpha^4 = 1 + 2\alpha + \alpha^2$.)
