# Prove S is a Group

• Dec 8th 2009, 06:59 PM
mesmo
Prove S is a Group
Hi again.

Well, I have another question.
Suppose A is a ring with unity denoted by 1. Let S be the set of elements in A that have multiplicative inverses in A.

I want to prove S is a group under mult. I know the four cases that must be proved. However, I'm having a little difficulty as to how I should prove that these particular things are in the set of elements S.

For example, the last case is to prove that for every element in S there is a multiplicative inverse. However, I am having a hard time thinking as to how I should prove the the inverses reside in S.

If this is vague, I am more than happy to explain what I mean.
• Dec 9th 2009, 02:00 AM
Swlabr
Quote:

Originally Posted by mesmo
Hi again.

Well, I have another question.
Suppose A is a ring with unity denoted by 1. Let S be the set of elements in A that have multiplicative inverses in A.

I want to prove S is a group under mult. I know the four cases that must be proved. However, I'm having a little difficulty as to how I should prove that these particular things are in the set of elements S.

For example, the last case is to prove that for every element in S there is a multiplicative inverse. However, I am having a hard time thinking as to how I should prove the the inverses reside in S.

If this is vague, I am more than happy to explain what I mean.

You have 4 things to show: associativity, identity, closure, and inverses.

Associativity holds as your multiplication is just the multiplication in the ring, which is associative.

Identity holds as $1 \in S$.

So closure and inverses remain to be proven. However, if we prove closure we have proven inverses (why?).

So, $x \in S$, $y \in S$. This means we can find $x^{\prime}$, $y^{\prime}$ such that $xx^{\prime}=1$ and $yy^{\prime}=1$. Then can you think up an inverse for $xy$?
• Dec 9th 2009, 05:48 AM
mesmo
Thanks Swlabr....i think i am understanding.

Okay, because this is a ring with unity, it automatically follows that this unity has to be an element of S? (same thing with associative multiplication?) I was thinking along those lines but I kept seeing S as something detached from the ring.

xx' = 1 and yy' = 1. xx' = yy'....? Is this going the write way? Since they are both equal to one is it correct to equate them to each other? however, this is as far as I've gotten. Is closure the same way? Thanks for the help so far.
• Dec 9th 2009, 05:56 AM
Swlabr
Quote:

Originally Posted by mesmo
Thanks Swlabr....i think i am understanding.

Okay, because this is a ring with unity, it automatically follows that this unity has to be an element of S? (same thing with associative multiplication?) I was thinking along those lines but I kept seeing S as something detached from the ring.

xx' = 1 and yy' = 1. xx' = yy'....? Is this going the write way? Since they are both equal to one is it correct to equate them to each other? however, this is as far as I've gotten. Is closure the same way? Thanks for the help so far.

You want an element $r \in R$ such that $(xy)r=1$. We know there is an element which gives us $xyr_1 = x$. What is this element $r_1$? Can you think of an element $r_1 \in R$ such that $xyr_1r_2=xr_1=1$? Set $r=r_1r_2$ and you are done!

I tend to think of this set " $S$" as everything close to the identity.