let G be a group with operation . and let p,q be two elements of G. simplify p.q.q.q^-1.p^-1.q^-1 as much as possible.does answer change if G is ABELIAN?
p.q.(q.q^-1).p^-1.q^-1
p.q.p^-1.q^-1
If Abelian, the p.q=q.p and we can go on:
q.(p.p^-1).q^-1
q.q^-1
1
thanks steve SO DOES ANSWER CHANGE IF ABELIAN?
To simplify the expression p.q.q.q^-1.p^-1.q^-1, we will apply the group operation to the elements in the given order.
First, we have:
p.q = (pq)
Next, we have:
(pq).q = (pq)q = pq^2
Then, q^-1 is the inverse of q. So,
pq^2.q^-1 = pq^(2-1) = pq
After that, p^-1 is the inverse of p. So,
(pq).p^-1 = (pq)p^-1 = pqp^-1 = pp^-1 = e (the identity element of the group)
Finally, q^-1 is the inverse of q. So,
e.q^-1 = eq^-1 = q^-1
Therefore, p.q.q.q^-1.p^-1.q^-1 simplifies to q^-1 in a non-Abelian group.
Now, if G is Abelian, it means that the group operation is commutative, i.e., for any elements p and q in G, pq = qp.
In an Abelian group, we can rewrite the expression p.q.q.q^-1.p^-1.q^-1 as follows:
p.q.q.q^-1.p^-1.q^-1 = pp^-1.q.q.q^-1.q^-1 = eq.q.q.q^-1.q^-1 = q.q^-1.q^-1 = qq^-1.q^-1 = e.q^-1 = q^-1
So, even in an Abelian group, the answer remains the same, which is q^-1.