# Why a set is not a group?

1. Nov 6, 2009

### hitmeoff

1. The problem statement, all variables and given/known data
Why does the set {$$\overline{1}$$,$$\overline{2}$$,$$\overline{3}$$} of nonzero classes modulo 4 fail to be a group under addition? Under multiplication?

2. Relevant equations

3. The attempt at a solution

Under addition it fails because there is no identity element such that e+s=s for s $$\epsilon$$ S. (ie 0)
And because theres no additive inverse (ie -1, -2, -3) such that -a + a = e (0)

Right?

And under multiplication if fails because theres no multiplicative inverse such that a-1$$\cdot$$a=e

Right?

2. Nov 6, 2009

### lanedance

be a little careful, you need to show/check what you're stating, which in the examples you've given means testing alot of cases.. a counter example may be easier...

for the identity element, this works, but you need to pick an element & test all otehrs to show they're not identities
for the additive inverse, note 3+2 = 1, so some additive inverses exist
for a counter example, is it closed under addition...?

for the multiplticative group
1*1 = 1
3*3 = 1
so some inverses exist... so that argument isn't solid, but can be used for 2
once again, might be worth checking closure...

3. Nov 6, 2009

### lanedance

edited & updated above for clarity

4. Nov 6, 2009

### hitmeoff

Ahhhh, I see.

I still have to do a mod 4 after performing the addition and multiplication operations. I knew I was wrong but wasn't sure why. Thanks.

5. Nov 6, 2009

### hitmeoff

So under addition: 1+3 = 0 and zero is not in the set, not closed
Under multiplication 2x2 = 0, not closed

6. Nov 6, 2009

Correct