Ha-ha, you forgot to translate w into English!
Wikipedia says that associativity holds. In the worst case, you have 8 cases to check.
You mean f * w = f.
The rest seems OK.
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
Ha-ha, you forgot to translate w into English!
Wikipedia says that associativity holds. In the worst case, you have 8 cases to check.
You mean f * w = f.
The rest seems OK.
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 .