It is well-known that if $i\in\mathbb Z$, then any group $G$ is abelian provided that $(ab)^k=a^kb^k$ holds for $k=i,i+1,i+2$ and for all $a,b\in G$ (see for example this question). Are there other $3$-subsets $K$ of $\mathbb Z$ such that, if $G$ is any group satisfying $(ab)^k=a^kb^k$ for every $a,b\in G$ and for all $k\in K$, then the group $G$ is abelian? More generally, are all these subsets known?
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Interesting question! – Nicky Hekster Oct 04 '13 at 09:23
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1Any $K$ containing $2$ will do. I wonder if you have two distinct integers $k, l > 2$, is it possible to find a nonabelian group satisfying $(ab)^k = a^kb^k$ and $(ab)^l = a^lb^l$? – spin Oct 04 '13 at 11:25
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@spin : I dont think so, take for instance the integers $3$ and $5$. – Yassine Guerboussa Oct 04 '13 at 18:21
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A group with the property $(ab)^k=a^kb^k$ is termed $k$-abelian, These groups are characterized by J. Alperin in http://cms.math.ca/cjm/v21/cjm1969v21.1238-1244.pdf.
It is straightforward to see that being $k$-abelian is equivalent to being $1-k$-abelian.
Also, If $G$ is $k$-abelian then the center quotient $G/Z(G)$ has a finite exponent dividing $k(k-1)$.
Assume the identity $(ab)^i=a^ib^i$ holds in a group $G$, for $i=k,n$ and that $(k(k-1),n(n-1))=2$, Then $G$ is abelian.
Indeed, the exponent of $G/Z(G)$ divides $k(k-1)$ and $n(n-1)$, thus $G/Z(G)$ has exponent at most 2, so it is abelian. It follows that $G$ has class at most $2$.
Now let $a,b \in G$, one has $(ab)^i=a^ib^i[b,a]^{i(i-1)/2}=a^ib^i$, for $i=k,n$. It follows that $[b,a]^{i(i-1)/2}=1$, for $i=k,n$, therefore $[b,a]=1$. Thus $G$ is abelian.
Yassine Guerboussa
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@Matemáticos Chibchas: your claim holds at least for any triplet, for which two components satisfy the above condition. – Yassine Guerboussa Oct 04 '13 at 18:25