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In a group $(Z,*)$, $(a*b)^{5}=a^{5}*b^{5},\forall a,b\in Z$ and $(a*b)^{3}=a^{3}*b^{3}$ then prove that $Z$ is abelian. I know that for three consecutive integer if $(a*b)^{i}=a^{i}*b^{i},\forall a,b\in Z$ holds then $Z$ is abelian. I know i have to use this property and i have to use three consecutive integer $3,4$ and $5$. But i am stuck.

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I take it that $Z$ is supposed to be $G$. If so, please edit accordingly. – Gerry Myerson Apr 17 '12 at 3:43
Yes in this case $G=Z$. – Kns Apr 17 '12 at 3:45
Then edit the question to reflect that! – Mariano Suárez-Alvarez Apr 17 '12 at 3:46
up vote 23 down vote accepted

(I'll write $xy$ instead of $x*y$ for simplicity.)

$(ab)^5=a^5b^5\implies ababababab=aaaaabbbbb\implies babababa=aaaabbbb$

$(ab)^3=a^3b^3\implies ababab=aaabbb\implies baba=aabb$

From $baba=aabb$, we get $babababa=aabbaabb$. Combining with $babababa=aaaabbbb$ gives $aabbaabb=aaaabbbb$, so $bbaa=aabb$. Combining again with $baba=aabb$ gives $bbaa=baba$, so $ba=ab$. This is true for any $a,b\in Z$, so $Z$ is abelian.

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A group $G$ is said to be $n$-abelian if $(ab)^n = a^nb^n$ for all $a,b\in G$. Given a group $G$, the "exponent semigroup of $G$" is defined to be $$\mathcal{E}(G) = \{n\in\mathbb{Z}\mid (ab)^n =a^nb^n\text{ for all }a,b\in G\}.$$ Properties of $\mathcal{E}(G)$ established by Levi (see this previous answer) imply that if $3,5\in\mathcal{E}(G)$, then $\mathcal{E}(G)=\mathbb{Z}$, giving your desired conclusion.

For a slightly less high-powered proof, note that in general we have that $$(ab)^n = a^nb^n \iff (ba)^{n-1}=a^{n-1}b^{n-1}$$ which can be obtained by cancelling a single $a$ and a single $b$. Therefore, if $n\in\mathcal{E}(G)$, then $$(xy)^{(n-1)^2} = \bigl( (xy)^{n-1}\bigr)^{n-1} = \bigl( y^{n-1}x^{n-1}\bigr)^{n-1} = (x^{n-1})^{n-1}(y^{n-1})^{n-1} = x^{(n-1)^2}y^{(n-1)^2}.$$ That is, $n\in\mathcal{E}(G)$ implies $(n-1)^2\in\mathcal{E}(G)$.

Here, you know $3\in\mathcal{E}(G)$, hence $(3-1)^2 = 4\in\mathcal{E}(G)$. So now we have that $3$, $4$, and $5$ are all in $\mathcal{E}(G)$, and as you note, it is known that if $\mathcal{E}(G)$ contains three consecutive integers then it contains all integers, hence $G$ is abelian.

(This is essentially jgnr's answer, only cast a bit more generally)

As an alternative, but also somewhat high-powered proof, we can follow Steve Ds good observation: by the theorem of Alperin quoted in the answered linked to above, a $3$-abelian group is a quotient of a subgroup of a direct product of abelian groups, groups of exponent $3$, and groups of exponent $2$; but the latter are abelian as well, so $G$ will be a quotient of a subgroup of a direct product of abelian groups and groups of exponent $3$. If you had an element that is not central (that does not commute with everything) then it would have to be an element of order $3$. But by the same characterization, since $G$ is $5$-abelian it is a quotient of a subgroup of a direct product of abelian groups, groups of exponent $4$, and groups of exponent $5$. So a noncentral element would have to have order dividing $20$. Since $\gcd(3,20)=1$, there can be no noncentral elements (it would have to have exponent $3$ and exponent $20=4\times 5$, hence exponent $1$, hence be trivial, hence it is central). So $G=Z(G)$, hence $G$ is abelian.

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Surely Alperin's result is good enough here? 3-abelian implies the only non-central elements must have exponent 3, which are ruled out by 5-abelian. – user641 Apr 17 '12 at 4:23
@SteveD: Good point; yes. – Arturo Magidin Apr 17 '12 at 4:29

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