# How do I show that the unit group of $\mathbb{Z}\times\mathbb{Z}/5\mathbb{Z}$ is a cyclic group of order 10? [closed]

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 $$(a_1,\overline{b_1})\cdot(a_2,\overline{b_2})=(a_1a_2,\overline{a_1b_2+a_2b_1}).$$

## closed as off-topic by user26857, Tim Raczkowski, graydad, user223391, user91500Nov 16 '15 at 3:55

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user26857, Tim Raczkowski, graydad, Community, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.

• Really sorry that I forgot to put the edit in from the start :( – user235238 Oct 11 '15 at 16:49

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.

• I dont know which is a generator – user235238 Oct 11 '15 at 17:30
• @TonyStrong try all elements. A generator is one whose powers cover the whole group. If you try all elements, you ought to find one. I gave some restrictions to the set of elements you have to try. Go ahead and try! – MickG Oct 11 '15 at 18:18
• i think (-1,3) is one – user235238 Oct 11 '15 at 18:19
• @TonyStrong The square is $(1,-6)=(1,1)$. The cube is $(1,1)(-1,3)=(-1,2)$. Going on, $(-1,2)(-1,3)=(1,-5)=(1,2)$, $(1,2)(-1,3)=(-1,1)$, so the order is at least 6, but it must divide 10 by Lagrange's theorem, hence it is 10. So we are done. I will update the answer. – MickG Oct 11 '15 at 18:23

$R$ 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: $$U(\mathbb{Z}\times \mathbb{Z}/5)\cong U(\mathbb{Z})\times U(\mathbb{Z}/5)\cong \mathbb{Z}_2 \times \mathbb{Z}_4.$$

• please see the edit – user235238 Oct 11 '15 at 16:37