# Thread: Proof for (Abelian) Group - or not?

1. ## Proof for (Abelian) Group - or not?

Hello together!

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

$G = \{w, f\}; a*b := (a \leftrightarrow b)$

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

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

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

Originally Posted by MageKnight
2.) Associativity
???
Wikipedia says that associativity holds. In the worst case, you have 8 cases to check.

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.

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

I am not sure I understand your definition for $a \leftrightarrow b$. To me, that reads $a$ if and only if $b$. 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 $a*(b*c) = (a*b)*c$ for all $a,b,c\in G$. So, try it out. There are only eight possible combinations for $a,b,c \in G$.