abstract algebra - Group of units in the rings $mathbb I_9$ and $mathbb I_{15}$?

The question I need help is:

Prove that $U(\mathbb I_9) \cong \mathbb I_6$ and $U(\mathbb I_{15}) \cong \mathbb I_4 \times \mathbb I_2$.

U() is the group of units in a ring

All the "I" are the integers mod the subscript.

I'm really confused because we just now started learning about commutative rings but have never once mentioned any of the things in this problem. I don't know what a group of units is and the like. I'm lost :(

Answer

I don't know what a group of units is and the like.

Unit (also called invertible element) of a ring $R$ is an element $a\in R$ for which there exists exists $b\in R$ such that $ab=1$. I.e., it is an element which has a multiplicative inverse.

For example, if you work with $\mathbb I_9=\{0,1,\dots,8\}$, then:

• $2$ and $5$ are units, since $2\cdot 5=1$.


• $3$ is not a unit, since by multiplying $3a$, $a\in\mathbb I_9$, you can only get numbers $0$, $3$ and $6$.


• Can you check whether $4$ is a unit. (I.e., can you find $b\in\mathbb I_9$ such that $4a=1$?

If you do everything correctly, you should find out that units are $U(\mathbb I_9)=\{1,2,4,5,7,8\}$.

Now you can check that that this set is closed under multiplication. And that it is a group. (Or maybe you have learned this as a theorem on your lecture.)

Now $\mathbb I_6$ is a cyclic group. To show that $\mathbb I_6$ is isomorphic to $U(\mathbb I_9)$ it suffices to find a generator of $(U(\mathbb I_9),\cdot)$. Can you find an element such that $U(\mathbb I_9)=\{1,a, a^2, a^3, a^4, a^5\}$? If you find such an element, then $1\mapsto a$ uniquely determines an isomorphism $(\mathbb I_6,+) \to (U(\mathbb I_9),\cdot)$.