Why Is {\overline{1},\overline{2},\overline{3}} Modulo 4 Not a Group?

[hitmeoff]

Homework Statement

Why does the set {$$\overline{1}$$,$$\overline{2}$$,$$\overline{3}$$} of nonzero classes modulo 4 fail to be a group under addition? Under multiplication?

Homework Equations

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 there's no additive inverse (ie -1, -2, -3) such that -a + a = e (0)

Right?

And under multiplication if fails because there's no multiplicative inverse such that a^-1$$\cdot$$a=e

Right?

 

[lanedance]

be a little careful, you need to show/check what you're stating, which in the examples you've given means testing a lot 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...

 

[lanedance]

edited & updated above for clarity

 

[hitmeoff]

be a little careful, you need to show/check what you;'e stating

for the additive inverse, note 3+2 = 1, so some additive inverses exist
is it closed under addition though?

for the multiplticative group
1*1 = 1
3*3 = 1
so some inverses exist...
once again, might be worth checking closure...

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.

 

[hitmeoff]

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

 

[VeeEight]

Correct