1. ## Group of Units

Show that the set of units, U, of R forms a group under multiplication.

My thoughts... U is a group since if we multiply a unit by a unit we get a unit. (not sure if true and would probably need a proof...)
Formally, let a,b be units. Then a|1 and b|1. So ab|1, hence ab is a unit. So is a group under multiplication...

Just a draft of what i think the answer is. Pretty sure there's more to it though. Am i on the right track?

Show that the set of units, U, of R forms a group under multiplication.

My thoughts... U is a group since if we multiply a unit by a unit we get a unit. (not sure if true and would probably need a proof...)
Formally, let a,b be units. Then a|1 and b|1. So ab|1, hence ab is a unit. So is a group under multiplication...

Just a draft of what i think the answer is. Pretty sure there's more to it though. Am i on the right track?
1 is clearly a unit since $\displaystyle 1\cdot1=1=1\cdot1.$
If $\displaystyle a$ and $\displaystyle b$ are units, then $\displaystyle aa'=1=a'a$ and $\displaystyle bb'=1=b'b$ for some $\displaystyle a',b'\in R.$ Hence $\displaystyle (ab)(b'a')=1=(b'a')(ab)$ showing that $\displaystyle ab$ is a unit.
And the above also shows that the multiplicative inverse of $\displaystyle a$ is $\displaystyle a'.$ Hence the set of all units of a ring with unity forms a group under multiplication.
Notice that I have been careful not to assume that $\displaystyle R$ is a commutative ring since you did not state whether it is or not.