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 ?
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Possible duplicate of Let $G$ be a group, where $(ab)^3=a^3b^3$ and $(ab)^5=a^5b^5$. Prove that $G $ is an abelian group. – Dietrich Burde Dec 25 '17 at 20:12
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1Possible duplicate of Let $G$ be a group, where $(ab)^3=a^3b^3$ and $(ab)^5=a^5b^5$. Prove that $G $ is an abelian group. – Dec 26 '17 at 15:19
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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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+John Watson thank you but please can you explain (*) to me that the main point to me .. – Taha Guma El Turki Topology Dec 25 '17 at 10:04
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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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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
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@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
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@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
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DonAntonio, I edit my answer. I want to thank you for polite and humble comment. – 1ENİGMA1 Dec 26 '17 at 18:54