# Problem from Herstein on group theory

The problem is:

If $G$ is a finite group with order not divisible by $3$, and $(ab)^{3} = a^{3} b^{3}$ for all $a,b \in G$, then show that $G$ is abelian.

I have been trying this for a long time but not been able to make any progress. The only thing that I can think of is: $$ab \cdot ab \cdot ab = aaa \cdot bbb \implies (ba)^{2} = a^{2} b^{2} = aabb = (\text{TPT}) abba.$$ Now, how can I prove the last equality? If I write $aabb = ab b^{-1} abb$, then in order for the hypothesis to be correct, $b^{-1} abb = ba \implies ab^{2} = b^{2} a$. Where am I going wrong? What should I do?

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• Suppose that $x \in G$ satisfies $x^{3} = e$. Then we cannot have $x \neq e$, otherwise $x$ would have order $3$, which implies that $3$ divides $|G|$ (recall that the order of a group element divides the order of the group). Hence, $$\forall x \in G: \quad x^{3} = e ~ \Longrightarrow ~ x = e,$$ and as $(ab)^{3} = a^{3} b^{3}$ for all $a,b \in G$, we see that the function $\phi: G \to G$ defined by $$\forall x \in G: \quad \phi(x) \stackrel{\text{def}}{=} x^{3}$$ is an injective group homomorphism.

• Now, $$\forall a,b \in G: \quad ababab = (ab)^{3} = a^{3} b^{3} = aaabbb.$$ Hence, $$\forall a,b \in G: \quad baba = aabb, \quad \text{or equivalently}, \quad (ba)^{2} = a^{2} b^{2}.$$ Using this fact, we obtain \begin{align} \forall a,b \in G: \quad (ab)^{4} &= [(ab)^{2}]^{2} \\ &= [b^{2} a^{2}]^{2} \\ &= (a^{2})^{2} (b^{2})^{2} \\ &= a^{4} b^{4} \\ &= aaaabbbb. \end{align}

• On the other hand, \begin{align} \forall a,b \in G: \quad (ab)^{4} &= abababab \\ &= a (ba)^{3} b \\ &= a b^{3} a^{3} b \\ &= abbbaaab. \end{align}

• Hence, for all $a,b \in G$, we have $aaaabbbb = abbbaaab$, which yields $$\phi(ab) = a^{3} b^{3} = b^{3} a^{3} = \phi(ba).$$ As $\phi$ is injective, we conclude that $ab = ba$ for all $a,b \in G$.

Conclusion: $G$ is an abelian group.

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Thanks a lot for the solution. But how can you say that $\phi(x)=x^3$ is injective? –  user23238 Jan 21 '13 at 6:44
What was your motivation for considering $(ab)^4$? It is a bit arbitrary for me. –  user23238 Jan 21 '13 at 6:46
Got it! $x^3=y^3 \implies x^3(y^{-1})^3\implies (xy^{-1})^3=e\implies x=y$ –  user23238 Jan 21 '13 at 6:59
@ramanujan_dirac: Hi Ramanujan. There is an even easier way to show that $\phi$ is injective. First use the property $\forall a,b \in G: ~ (ab)^{3} = a^{3} b^{3}$ to prove that $\phi$ is a group homomorphism. Then use the fact that $\forall x \in G: ~ x^{3} = e \Longrightarrow x = e$ to deduce that $\ker(\phi) = \{ e \}$. It follows immediately that $\phi$ is injective. My consideration of $(ab)^{4}$ was an attempt to play around with identities to see what I could get. My goal was to somehow obtain $\forall a,b \in G: ~ (ab)^{3} = (ba)^{3}$. –  Haskell Curry Jan 21 '13 at 20:53

Hints (remember: $\,|G|<\infty\,\,\,and\,\,\,3\,\nmid\, |G|\,$): $\,\,\forall\,\,a,b\in G\,$

\begin{align*}(1)&\;\;\;\text{Show that}\,\,\,\,(ba)^2=a^2b^2\\{}\\(2)&\;\;\;\text{Prove that}\;\;f:G\to G\,\,\,,\,\,f(x):=x^3\,\,,\,\,\text{is an isomorphism}\\{}\\(3)&\;\;\;\text{Define}\,\,z:=\left(aba^{-1}\right)^3 \longrightarrow \begin{cases}z=ab^3a^{-1},\;\;\;\text{and also}\\{}\\z=f(a)f(b)f(a^{-1})=a^3b^3a^{-3}\end{cases}\\{}\\(4)&\;\;\;\text{Using(2)-(3) , show that}\;\;a^2\in Z(G)\Longleftrightarrow a^2g=ga^2\,\,,\,\forall\,g\in G\\{}\\(5)&\;\;\;\text{Finally, use (1) to show that}\,\,\,ab=ba\end{align*}

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For (3,(4)) you could use: $(ab)^2=b^2a^2\Rightarrow a^3b^3=(ab)^3=(ab)(ab)^2=ab^3a^2\Rightarrow a^2b^3=b^3a^2$. –  P.. Jan 20 '13 at 10:05
Good one, @Pambos: easier your way. +1 –  DonAntonio Jan 20 '13 at 10:12
@DonAntonio: I haven't been able to prove $a^2g=ga^2$, From 2, 3, and also by Pambos' comment I have proved that $a^2b^3=b^3a^2$, but how can we substitute $b^3$ by $g$. Aren't we assuming that any element can be written as the cube of another? Thanks. –  user23238 Jan 22 '13 at 2:20
Well, since by (2) we know that $\,f\,$ is surjective, then $\,\forall\,x\in G\,\,\exists\,g_x\in G\,\,s.t.\,\,x=g_x^3\,$ , so by what you've proved $\,a^2x=xa^2\,\,\forall\,\,x\in G\Longleftrightarrow a^2\in Z(G)\,$ ... –  DonAntonio Jan 22 '13 at 2:44

The way, I am writing here, is from my old notes and personally I prefer the other approaches. But, maybe the given additional points below, inspire you for other problems like this problem.

We can prove that if for an integer $n$ and every $a,b\in G$, $(ab)^n=a^nb^n$, then $$(aba^{-1}b^{-1})^{n(n-1)}=e$$ The proff is easy. In fact, $$(aba^{-1}b^{-1})^{n^2}=[(aba^{-1}b^{-1})^n]^n=[a^n(ba^{-1}b^{-1})^n]^n=...=a^nb^na^{-n}b^{-n}\\\ (aba^{-1}b^{-1})^{n}=(ab)^n(a^{-1}b^{-1})^n=a^nb^na^{-n}b^{-n}$$

In your problem, we assume $G$ is not abelian, so there exist $a,b\in G, aba^{-1}b^{-1}\neq e$. According to above recently fact $$(aba^{-1}b^{-1})^6=e$$ since we know $(ab)^3=a^3b^3$. So $|aba^{-1}b^{-1}|\big| 6$ and because of $3\nmid|G|$ so $|aba^{-1}b^{-1}|=2$. This means that $(aba^{-1}b^{-1})^2=e$. On the other hand, $$(ab)^3=a^3b^3\Longrightarrow (ba)^2=a^2b^2$$ (see @Haskell's answer) then $(a^{-1}b^{-1})^2(ab)^2=e$ or $(ab)^2=(ba)^2=a^2b^2$ or $ab=ba$. A nice contradiction!

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+1 nice contribution (variety is always good)! –  amWhy Feb 13 '13 at 0:06