Proof: (R,+) is a Group but (R,*) is Not

[fk378]

Homework Statement

Let (R,+,*) be a commutative ring with identity. Show that (R,+) is a group but (R,*) is never a group.

The Attempt at a Solution

This question confuses me because I thought a group was defined for a set with a binary operation, i.e. a set that uses multiplication.

Also, does binary operation imply multiplication and addition are defined?

 

[tiny-tim]

Homework Statement

Let (R,+,*) be a commutative ring with identity. Show that (R,+) is a group but (R,*) is never a group.

The Attempt at a Solution

This question confuses me because I thought a group was defined for a set with a binary operation, i.e. a set that uses multiplication.

Hi fk378!

"binary operation" means any operation on only two things …

nearly every operation we use is binary …

Also, does binary operation imply multiplication and addition are defined?

"binary operation" means one operation … and multiplication and addition are two operations … so nooo.

"binary operation" only means that a-operation-b is defined for all a and b (in that order).

It needn't be a group, or commutative, or anything … it could be totally random!

We can call it anything we like … we normally call an operation "multiplication", just because that's the way we write it … but that doesn't mean it has any of the properties of "usual" multiplication.

If a set has two binary operations, obviously we can't call both of them "multiplication".

Anyway … how would you show that (R,*) isn't a group? ​

 

[fk378]

A set is not a group if
1) it is not closed under the operation
2) not associative
3) no identity element e
4) no inverse.

But can't this also apply to (R,+)? How can (R,*) --never-- be a group?

 

[tiny-tim]

But can't this also apply to (R,+)? How can (R,*) --never-- be a group?

Hi fk378!

Yes … and that's what the question asks you to prove!

How can (R,*) --never-- be a group?

i] what is the identity under * ?

ii] does every element of R have an inverse under * ?

 

[fk378]

True. I see that 0 does not have an inverse. But can't you define the set to not include 0? [1, infinity)?

 

[Dick]

True. I see that 0 does not have an inverse. But can't you define the set to not include 0? [1, infinity)?

i) No, the set is (R,*). You can't just redefine it. ii) Even if you do, there still may be elements without inverses. Look (Z,+,*), the ring of integers. 2 doesn't have an inverse.

 

[fk378]

Oh I see. Thank you. Dick, for your example of (Z,+,*), does any number have an inverse (besides 1)?

 

[Dick]

-1 does. That's about it.