There is a cool problem from Herstein’s *Topics in Algebra*. It states

Let be a finite group whose order is not divisible by 3. Suppose that for all for all . Prove that must be abelian.

I recall solving this problem a year ago for an assignment given in intro to abstract algebra course. The assignment is somewhere on my shelf, probably covered in dust. But I remember the way I started attacking it. It was something about group homomorphisms. By letting , it follows from the hypothesis that . But then again, the group homomorphism is introduced later in the book, so there should be an easier way of solving it. Here is a different approach:

*Solution*: Let . We want to show that and commute. By hypothesis, and using the cancellation law, we obtain:

Similarly,

Combining these two results,

Using , we get

Now, write the above equation differently,

The last equality is the really the key here. Notice that we haven’t yet used the fact that order of is not divisible by 3. We shall use it in a moment. First observe that implies

where denotes the identity element of the group (I am using the same notation as Herstein). Then, our hypothesis implies that

But the order of is not divisible by 3, so Lagrange’s theorem implies that no element can satisfy (more precisely, if this were true for some , then would imply the cyclic subgroup generated by element has size 3 but that violates Lagrange’s theorem). Observe that we have shown . Therefore, or equivalently,

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