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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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3 Answers 3

up vote 5 down vote accepted
  • 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
1  
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

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