How do I show that the unit group $R^*$ of $R=\mathbb{Z}\times\mathbb{Z}/5\mathbb{Z}$ is a cyclic group of order 10? I am allowed to use the fact that $R$ is isomorphic with $\mathbb{Z}[X]/(5X,X^2)$. Also is $R$ a domain?
EDIT: $R$ is the additive group with product action $$(a_1,\overline{b_1})\cdot(a_2,\overline{b_2})=(a_1a_2,\overline{a_1b_2+a_2b_1}).$$
On
$R$ is not a domain, since $(1,0)\cdot (0,1)=(0,0)$, but neither $(1,0)$ nor $(0,1)$ are zero. What is $U(R\times S)$ in terms of $U(R)$ and $U(S)$ for rings $R$ and $S$ ? Do you know that $U(\mathbb{Z}/5)\cong \mathbb{Z}_4$ and $U(\mathbb{Z})\cong \mathbb{Z}_2$ ?
Answer: $$ U(\mathbb{Z}\times \mathbb{Z}/5)\cong U(\mathbb{Z})\times U(\mathbb{Z}/5)\cong \mathbb{Z}_2 \times \mathbb{Z}_4. $$
Well, first of all it seems clear to me that for $(a,b)$ to belong to the unit group, $a$ must be 1 or $-1$. So we are left with at most 10 elements, $\{\pm1\}\times\mathbb{Z}/5\mathbb{Z}$.
Let us now take $(1,\bar b)$. Then $(1,\bar b)(a,\bar c)=(a,\overline{c+ba})$, but $a=\pm1$, so for $(a,\bar c)$ to be the inverse of $(1,\bar b)$, we need either $c+b\equiv1\mod5$ or the same for $c-b$. Whatever $b\in\mathbb{Z}$ you take, such a $c$ will always exist: either $b$ or $-b$. Similarly, you can show $(-1,\bar b)$ has an inverse. So all elements in $\{\pm1\}\times\mathbb{Z}_5$ have an inverse w.r.t. this product.
The rest is just trying your luck and finding a generator. One thing is sure: it must be of the form $(-1,\bar b)$, or you will never get anything of that form in the generated subgroup. If my mental calculations are correct, $(-1,\bar2)$ is not a generator. I leave it to you to find one.
Update
As you suggested in a comment, $(-1,3)$ is a generator. Its first few powers are $(-1,3),(1,1),(-1,2),(1,2),(-1,1)$. The identity element is $(1,0)$, so none of these is the id. So the order is at least 6. But the order must divide 10, which is the order of the group, by Lagrange's theorem. Hence we have a generator.