Interesting Exercise

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

Let G be a finite group whose order is not divisible by 3. Suppose that (ab)^3=a^3b^3 for all a,b\in G for all a,b\in G. Prove that G 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 \phi(x)=x^3, it follows from the hypothesis that \phi(ab)=\phi(a)\phi(b). 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 a,b\in G. We want to show that a and b commute. By hypothesis, and using the cancellation law, we obtain:

(ab)^3=a^3b^3 \ \ \Rightarrow \ \ a(ba)^2b=a^3b^3 \ \ \Rightarrow \ \ (ba)^2=a^2b^2

Similarly,

(ab)^9=((ab)^3)^3=(a^3b^3)^3=a^9b^9 \ \ \Rightarrow \ \ a(ba)^8b=a^9b^9 \ \ \Rightarrow \ \ (ba)^8=a^8b^8

Combining these two results,

(a^2b^2)^4=(ba)^8=a^8b^8 \ \ \Rightarrow \ \ a^2(b^2a^2)^3b^2=a^8b^8 \ \ \Rightarrow \ \ (b^2a^2)^3=a^6b^6

Using b^2a^2=(ab)^2, we get

(ab)^6=(b^2a^2)^3=a^6b^6 \ \ \Rightarrow \ \ a(ba)^5b=a^6b^6 \ \ \Rightarrow \ \ (ba)^5=a^5b^5

Now, write the above equation differently,

(ba)^2(ba)(ba)^2=a^5b^5 \ \ \Rightarrow \ \ a^2b^2(ba)a^2b^2=a^5b^5 \ \ \Rightarrow \ \ b^3a^3=a^3b^3

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

a^3b^3a^{-3}b^{-3}=e

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

(aba^{-1}b^{-1})^3=(ab)^3(a^{-1}b^{-1})^3=a^3b^3a^{-3}b^{-3}=e

But the order of G is not divisible by 3, so Lagrange’s theorem implies that no element x\ne e can satisfy x^3=e (more precisely, if this were true for some x\ne e, then x^3=e would imply the cyclic subgroup generated by element x has size 3 but that violates Lagrange’s theorem). Observe that we have shown (aba^{-1}b^{-1})^3=e. Therefore, aba^{-1}b^{-1}=e or equivalently, ab=ba \blacksquare

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