## 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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