Another one in my series of Yahoo Answers this time using elementary properties of Groups.

**Problem**: Let be elements of group such that where is the identity element of . Prove that

**Solution**: The problem is indeed trivial if is abelian, i.e. for all . One must be careful not to assume such properties when dealing with constructs in abstract algebra. Nevertheless, the solution below is quite simple and uses only associativity of Groups, i.e. for all .

Lemma: For any there is a unique such that .

Proof: The lemma can be proven in two steps. First, show that such exists, and second, show that it must be unique.

Let . Since and is a group, it follows from the existence ofÂ inverses that and from group closure . Thus, (and the first step is done).

Let be such that , take to be the identity element of . Then

Thus it follows that if then it must be that , finishing the second step, hence proving out lemma.

Now we go back to our proof. Suppose

is the identity in , i.e.

Using group associativity we get

“Multiplying” by (on right) we get

Now multiplying by (on left) we get

Now from we have

and from

Combining and

Since is the identity element of , by definition

Finally, using our Lemma and with it follows that

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