Proof for (Abelian) Group - or not?

Hello together!

I need to show whether the following is a(n abelian) group or not:

I try to show the axioms.

1) Closure

Is given, since

w*f = f

f*w = f

w*w = w

f*f = f

2.) Associativity

???

3.) Identity

I need an element e with e*a = a*e = a. (for all elements of G)

So..

a*e = a <-> e = w.

because w*w = w and f*f = f.

4.) Inverse element;

I need an element i with

a*i = i*a = e.

the inverse for w is w, since w*w=w=e.

the inverse for f is f, since f*f=w=e.

5.) Commutativity:

Is a*b=b*a ?

This can be shown be the few examples:

w*f = f

f*w = f

Re: Proof for (Abelian) Group - or not?

Ha-ha, you forgot to translate w into English!

Quote:

Originally Posted by

**MageKnight** 2.) Associativity

???

Wikipedia says that associativity holds. In the worst case, you have 8 cases to check.

Quote:

Originally Posted by

**MageKnight** 3.) Identity

I need an element e with e*a = a*e = a. (for all elements of G)

So..

a*e = a <-> e = w.

because w*w = w and f*f = f.

You mean f * w = f.

The rest seems OK.

Re: Proof for (Abelian) Group - or not?

I am not sure I understand your definition for . To me, that reads if and only if . But, the multiplication you give in part (1) suggests otherwise. Then, for part (4), your multiplication changes: f*f = w in part (4) but f*f = f in part (1) and part (3). I think you are correct in part (4) since that multiplication is correct for the biconditional.

For part (3), you say w*w = w and f*f=f. I think you want to show w*w = w and f*w = w*f = f. So, w is the identity element.

For part (4), I agree with you.

For part (5), you have f*f = f*f = w, f*w = w*f = f, and w*w = w. Those are the three cases.

Anyway, for associativity, you need to show that for all . So, try it out. There are only eight possible combinations for .