Proving that the set of units of a ring is a cyclic group of order 4 The set of units of $\mathbb{Z}/10\mathbb{Z}$ is $\{\bar{1}, \bar{3}, \bar{7}, \bar{9}\}$, how can I show that this group is cyclic?
My guess is that we need to show that the group can be generated by some element in the set, do I need to show that powers of some element can generate all elements in the other congruence classes?
For example $7^2 = 49 \equiv 9 \pmod{10}$, i.e using $7$ we can generate an element in the congruence class of $9$, but can not generate $29$ for example from any power of $7$, so is it sufficent to say that an element is a generator if it generates at least one element in all other congruence classes?
 A: You had just the right idea, then got sidetracked. What you've seen is that $\overline 7^2=\overline 9.$ From this, it does, indeed, follow that $\overline 7^2=\overline{29},$ since we're thinking modulo $10.$
From there, $$7^3\equiv 9\cdot 7=63\equiv 3,$$ so that $\overline 7^3=\overline 3,$ and similarly, $\overline 7^4=\overline 1.$
As an alternative, we could show that $\overline 3^2=\overline 9,$ $\overline 3^3=\overline 7,$ and that $\overline 3^4=\overline 1,$ which is arguably even simpler, since $3^2=9,$ $3^3=27,$ and $3^4=81$ are math facts you probably know.
A: 
My guess is that we need to show that the group can be generated by some element in the set, do I need to show that powers of some element can generate all elements in the other congruence classes?

It is enough if you find one element such that its powers cover exactly the group. I would give $\bar 3$ a try.

For example $7^2 = 49 \equiv 9 \bmod 10$, i.e. using $7$ we can generate an element in the congruence class of $9$, but can not generate $29$ for example from any power of $7$, so is it sufficent to say that an element is a generator if it generates at least one element in all other congruence classes?

Edited following Cameron Buie's comment: $\bar 7$ actually works, as it generates $\bar 9$, $\bar 3$ and $\bar 1$, in that order.
Let's see what happens with $\bar 3$:

*

*$\bar 3^1 = \bar 3$

*$\bar 3^2 = \bar 9$

*$\bar 3^3 = \bar{27} = \bar 7$

*$\bar 3^4 = \bar{81} = \bar 1$
This is enough. On the one hand, we have covered all the elements in the subgroup. On the other hand, we are now satisfied that $\bar 3^4$ is the neutral element of the group, so we are positive that for any natural number $n$, $\bar 3^n = \bar 3^{n \bmod 4}$.
