# Set of units forming a group

1. Oct 19, 2004

### dogma

Hello out there! I'm a new guy here, so don't pick on me too much...I cry easy

I want to show that a set of units in a ring forms a group under multiplication. What steps would I take to show this?

Things that my feable brain knows:

1) if $$a$$ is a unit, it is invertible and $$a^-^1$$ is also a unit.
2) the product of units is a unit.
3) and that $$(ab)^-^1 = b^-^1 a^-^1$$

How should I proceed?

Thanks in advance for any and all help.

Best!

2. Oct 19, 2004

### matt grime

what does it mean for something to be a unit? now, write out the definitions of a group and by some basic algebraic manipulations show they are satisfied.

This is a reasonably elementary question, and I suspect that you've not thought long enough about it, that's all, especially as 1-2 above imply that it is a group directly.

3. Oct 19, 2004

### dogma

ah, the light bulb is now lit (just needed a little help finding that switch).

Thanks for enlightening me (sorry for the bad pun).

Take care.

4. Oct 19, 2004

### mathwonk

basically just be sure you always read and understand the definition of what you are trying to prove. I.e. a group is a set closed under an associative (binary) operation with an identity, and where every element has an inverse.

In a ring you have an associative multiplication with an identity, and the units are the elements that have multiplicative inverses, so this proof becomes easy once you know what the words mean.

Moral: before trying to prove every A is a B, always review the definitions of A and B first, as Matt advised.

Oh I guess you have to prove closure, i.e. the rpoduct of two units is a unit, which us your 3), so you had already done the only non trivial part of the proof.

Last edited: Oct 19, 2004