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I've been working on this problem listed in Herstein's Topics in Algebra (Chapter 2.3, problem 4):

If $G$ is a group such that $(ab)^i = a^ib^i$ for three consecutive integers $i$ for all $a, b\in G$, show that $G$ is abelian.

I managed to prove it, but I'm not very happy with my result (I think there's a neater way to prove this). Anyway, I'm just looking to see if there's a different approach to this.

My approach:

Let $j=i+1, k=i+2$ for some $i\in \mathbb{Z}$.

Then we have that $(ab)^i = a^ib^i$, $(ab)^j = a^jb^j$ and $(ab)^k = a^kb^k$.

If $(ab)^k = a^kb^k$, then $a^jb^jab =a^jab^jb$.

We cancel on the left and right and we have $b^ja = ab^j$, that is $b^iba = ab^j$.
Multiply both sides by $a^i$ on the left and we get $a^ib^iba = a^jb^j$, so $(ab)^iba = (ab)^j$.
But that is $(ab)^iba = (ab)^iab$.

Cancelling on the left yields $ab=ba$, which holds for all $a,b \in G$, and therefore, $G$ is abelian.

Thanks!

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I think you actually have the most obvious way of doing this –  Jay Kopper May 24 '11 at 3:08
    
I think this is the only way. What motivates you that there could be an another proof of this. –  user9413 May 24 '11 at 3:09
    
I don't know if there's another way; I just wanted to check if there was a simpler way because the other exercises from this unit were relatively straightforward and this one took me some time, so maybe I was overlooking a simpler solution. –  Fernando Martin May 24 '11 at 3:15
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That's also the way I did it. It would be less confusing, though, if you wrote $i+1$ and $i+2$ in full. –  Yuval Filmus May 24 '11 at 3:24
    
The only touch up I'd add is to extend the sentence "Let $j=i+1$...for some $i \in \mathbb{Z}$", such that $(ab)^i = a^ib^i$, ..., and $(ab)^k = a^kb^k$. At that point, you can move on to "Given $(ab)^k = a^kb^k, then...[rest of your proof]. You want to state the premise of the problem in full. As your proof now stands, you simply assume there are three consecutive integers. And then you proceed to argue: Then [it follows that] (ab)^i = a^ib^i....and....and...But that doesn't "follow" from the existence of three consecutive integers, alone. Are you following me? –  amWhy May 24 '11 at 3:41

4 Answers 4

up vote 15 down vote accepted

A very similar solution, but maybe slightly shorter: again let $i$, $i+1$ and $i+2$ be the three consecutive integers that work for $a$ and $b$. From $a^{i+1} b^{i+1} = (ab)^{i+1} = (ab)(ab)^i = aba^i b^i$ we get $a^i b = b a^i$. The same proof with $i$ replaced by $i+1$ gives $a^{i+1} b = b a^{i+1}$. Now $a b = a b a^i a^{-i} = a a^i b a^{-i} = a^{i+1} b a^{-i} = b a^{i+1} a^{-i} = b a$.

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I was writing this as a comment, but it's going to be extensive; I suspect most of the commenters might be interested, so here goes as a community wiki.

A group $G$ is said to be $n$-abelian if $(ab)^n = a^nb^n$ for every $a,b\in G$.

Alperin proved (The classification of $n$-abelian groups, Canad. J. Math. 21 (1969) 1238-1244, MR0248204 (40 #1458)) that the variety of $n$-abelian groups is the join of the variety of abelian groups, the variety of groups of exponent $n$, and the variety of groups of exponent dividing $n-1$ (so a group is $n$-abelian if and only if it is a quotient of a subgroup of a direct product of groups that are either abelian, of exponent $n$, or of exponent dividing $n-1$).

More recently, Primož Moravec (Schur multipliers and power endomorphisms of groups. J. Algebra 308 (2007), no. 1, 12–25. MR2290906 (2007j:20047)) proved that a finite $p$-group is $n$-abelian if and only if $n\in p^{e+r}\mathbb{Z}\cup(1+p^{e+r})\mathbb{Z}$, where $p^e$ is the exponent of $G/Z(G)$ and $r$ is an integer that can be computed, called the "exponential rank of $G$".

A closely related class of groups are the $n$-Bell and the $n$-Levi groups. A group is $n$-Bell if and only if $[x^n,y]=[x,y^n]$ for all $x,y\in G$, where $[a,b]$ is the commutator of $a$ and $b$. A group is $n$-Levi if $[x^n,y] = [x,y]^n$ for all $x,y\in G$. (Note that since $[a,b]=[b,a]^{-1}$, $n$-Levi implies $n$-Bell). A result I noted in passing some years ago was that if a group is $i$-, $i+1$-, and $i+2$-Levi for some integer $i$, then it is $n$-Levi for all $n$ (and hence nilpotent of class at most $2$), very similar to the problem at hand here. Also, $2$-Levi and $(-1)$-Levi imply $n$-Levi for all $n$, just like $(-1)$-abelian and $2$-abelian imply abelian (i.e., $n$-abelian for all $n$).

A nice paper on this subject is L-C Kappe's On $n$-Levi groups. Arch. Math. (Basel) 47 (1986), no. 3, 198–210, MR 0861866 (88a:20048). For a group $G$, let $$\begin{align*} \mathcal{E}(G) &= \{n\mid (xy)^n = x^ny^n\text{ for all }x,y\in G\}\\ \mathcal{L}(G) &= \{n\mid [x^n,y]=[x,y]^n\text{ for all }x,y\in G\}\\ \mathcal{B}(G) &= \{n\mid [x^n,y] = [x,y^n]\text{ for all }x,y\in G\}, \end{align*}$$ called, respectively, the "exponent semigroup", the "Levi semigroup", and the "Bell semigroup" of $G$ (the sets are closed under multiplication, so they are multiplicative semigroups of the integers). Kappe proves that if $W$ is a set of integers, then the following are equivalent:

  1. $W=\mathcal{E}(G)$ for some group $G$.
  2. $W=\mathcal{B}(H)$ for some group $H$.
  3. $W=\mathcal{L}(K)$ for some group $K$.

She also gives conditions for a set of integers to be an exponent semigroup, reminding readers that F.W. Levi proved, in Notes on Group Theory VII. The idempotent residue classes and the mappings $\{m\}$. J. Indian Math. Soc. (N.S.) 9 (1945), 37-42, that if $M$ is the smallest positive integer in $\mathcal{E}(G)$ for which $G^M$ is abelian, then there exist pairwise relatively prime integers $q_i$, $M=q_1\cdots q_t$, $2\lt q_i$, such that $\mathcal{E}(G)$ consists precisely of the integers that are congruent to either $0$ or $1$ modulo $q_i$ for $i=1,\ldots,t$; and that if $q_1,\ldots,q_t$ are pairwise relatively prime integers, $2\lt q_i$ for each $i$, and $S$ is the set of all integers that are congruent to either $0$ or $1$ modulo $q_i$ for each $i$, then there exists a group $G$ such that $\mathcal{E}(G)=S$.

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thanks a lot for these references. We speak of Exponent semigroup, Levi semigroup, etc. What is the operation on these sets? –  ogerard May 24 '11 at 6:33
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@ogerard: Multiplication; for example, if $n$ and $m$ are in the exponent semigroup, then so is $nm$, since $(xy)^{nm} = ((xy)^n)^m = (x^ny^n)^m = (x^n)^m(y^n)^m = x^{nm}y^{nm}$. –  Arturo Magidin May 24 '11 at 15:37

Preliminary comment: There is ambiguity in the statement of the problem, an ambiguity that I am told traces back to the original. I interpret the problem to mean that there exist consecutive integers $i$ such that $(ab)^i=a^ib^i$ for all $a$ and $b$. But the informal language of the problem permits the weaker interpretation that for every $a$ and $b$, there exist three consecutive integers $i$ such that $(ab)^i=a^ib^i$.

I will prove the result under the stronger "uniform $i$" condition. The arguments of the OP and of Robert Israel, proved under weaker assumptions, are therefore clearly "better." The reason for the argument below is that my parsing of the wording interprets the condition as uniform. And the stronger assumption permits a more structured, symmetric approach.

Let our exponents be $n$, $n+1$, and $n+2$.

Note that by associativity, $$(ab)^{n+1}=a(ba)^nb.$$ Thus $$a^{n+1}b^{n+1}=(ab)^{n+1}=a(ba)^{n}b=ab^{n}a^{n}b.$$ Cancellation now gives $$a^nb^n=b^na^n$$ from which we conclude that $$(ab)^n=(ba)^n.$$

A similar calculation with $n+1$ and $n+2$ yields $$(ab)^{n+1}=(ba)^{n+1}$$ and now division gives the desired result.

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Careful: the way I read the problem, it's not clear that the $n$ for which $a^n b^n = (ab)^n$ are the same ones for which $b^n a^n = (ba)^n$. (I'm disappointed with Herstein for being sloppy with quantifiers here: my interpretation is "for all $a, b \in G$ there are three consecutive integers $i$ ...", not "there are three consecutive integers $i$ such that for all $a, b \in G$ ..."). –  Robert Israel May 24 '11 at 6:34
    
@Robert Israel: I do not have the book, so need to go by the OP. It would be a very unusual reading of the words to interpret them as meaning "for all $a$ and $b$, there exist consecutive integers". I think of my answer as being proved under the hypotheses of the problem, and of yours and the OP's as being proved under weaker hypotheses. –  André Nicolas May 24 '11 at 12:22
    
The OP wrote the problem exactly as it appears in Herstein. If Herstein wanted your interpretation, I think it would have been more natural to write "$(ab)^i = a^i b^i$ for all $a,b \in G$, for three consecutive integers $i$". –  Robert Israel May 24 '11 at 16:09
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In fact, the exercise may be restated as saying "Let $a$ and $b$ be fixed. If there exist three consecutive integers $i$ such that $(ab)^i = a^ib^i$, then $a$ and $b$ commute." Since a group is abelian if and only if every $2$-generated subgroup is abelian, the full problem follows from this. –  Arturo Magidin May 24 '11 at 19:47

HINT $\rm\ \ \ (a\:b)^n\: =\:\ b^{-1}\ (\ b\ a\ )^{\:n+1}\:\ a^{-1}$

$\rm\quad\quad\Rightarrow\quad\ \ a^n\:b^n\ =\:\ b^{-1}(b^{n+1}a^{n+1})\:a^{-1} =\ b^n\: a^n $

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