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Let G be a group in which $(a*b)^3=a^3 * b^3$ and $(a*b)^5=a^5*b^5$ for all $a,b \in G$. Show that G is an Abelian group ?

nonuser
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2 Answers2

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Since $$(ab)^3 = a^3b^3 \Longrightarrow (ba)^2 = a^2b^2;\;\;\;\;(*)$$

and $$(ab)^5 = a^5b^5 \Longrightarrow (ba)^4 = a^4b^4;\;\;\;\;$$

so $$ (a^2b^2)^2 = ((ba)^2)^2= a^4b^4\Longrightarrow b^2a^2 = a^2b^2$$

using $(*)$ we get $$b^2a^2=(ba)^2 \Longrightarrow ba = ab$$

nonuser
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First observe that $(ab)(ab)(ab)=a^3b^3\Rightarrow a^2b^2=(ba)(ba)=(ba)^2$

So,

$(ab)(ab)(ab)(ab)(ab)=a^5b^5$

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

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

$\Rightarrow$ $b^3(ab)(ab)=(ba)(ba)b^3$

$\Rightarrow$ $b^2(ab)(ab)=abab^3$

$\Rightarrow$ $b^2(aba)=abab^2$

$\Rightarrow$ $b(ba)(ba)=abab^2$

$\Rightarrow$ $ba^2b^2=abab^2$

$\Rightarrow$ $ba^2=aba$

$\Rightarrow$ $ba=ab$.

1ENİGMA1
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  • How do you justify your second $;\implies;$ ? – DonAntonio Dec 25 '17 at 09:53
  • multiply $a^{-3}$ and $b^{-3}$ by left respectivlly – 1ENİGMA1 Dec 25 '17 at 09:56
  • Does not work since, a priori, $b^{-3}a^2b^5$ is not the same as $a^2b^2$. – Did Dec 25 '17 at 09:58
  • @upvoters Please explain your vote. Did you read the post before voting? – Did Dec 25 '17 at 09:58
  • @1Enigma then I get $$a^{-3}a^3b^3(ab)(ab)=a^{-3}a^5b^5\implies b^3(ab)(ab)=-a^2b^5$$and if I further multiply this on the left by $;b^{-3};$ I get$$(ab)(ab)=b^{-3}a^2b^5\ldots?$$ How is the last expression equal to $;a^2b^2;$ ? Did I miss anything? – DonAntonio Dec 25 '17 at 09:59
  • @1ENİGMA1 I think it would be a good idea to delete this answer as it is basically incorrect...or at least unexplained. – DonAntonio Dec 25 '17 at 10:14
  • DonAntonio, I edit my answer. I want to thank you for polite and humble comment. – 1ENİGMA1 Dec 26 '17 at 18:54