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The problem is to describe (up to isomorphism) all noncommutative groups $G$ containing two elements $a,b$ such that any element $g\in G$ can be uniquely written $g=a^ib^j$ for some $i,j \in {\mathbb Z}$. An obvious example is a semi-direct product of $\mathbb Z$ by itself. Are there other solutions ?

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What about the free product $\,\Bbb Z *\Bbb Z\,$, which is nothing else that the free group in two generators...? –  DonAntonio Sep 21 '12 at 18:46
    
@DonAntonio: can't write everything as $a^ib^j$ in that case –  Ben Millwood Sep 21 '12 at 18:48
    
Yes, I just noticed that: it's just one single element of the form $\,a^ib^j\,$ and not a (finite) product... –  DonAntonio Sep 21 '12 at 18:50
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@Mustafa: what are the two elements $a, b$ in this case? –  Qiaochu Yuan Sep 21 '12 at 18:59
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It is not true in general that $aba^{-1} = b^j$ implies that $a^{-1}ba$ is a power of $b$. It may do in this situation, but it does need justifying. –  Derek Holt Sep 22 '12 at 18:51

2 Answers 2

Still not a complete answer. By hypothesis we can write $aba^{-1} = a^i b^j$ and $a^{-1} ba = a^k b^{\ell}$ for unique $i, j, k, \ell$. Composing these operations in both orders gives

$$b = a^{i + jk} b^{j \ell} = a^{k + i \ell} b^{j \ell}$$

which by uniqueness gives $j \ell = 1, i + jk = k + i \ell = 0$.

Case: $j = \ell = 1$. Then $i + k = k + i = 0$, so $k = -i$. This gives

$$aba^{-1} = a^i b \Leftrightarrow b = a^{i-1} ba \Leftrightarrow bab^{-1} = a^{1-i}.$$

But we also have $b^{-1} ab = a^m b^n$ for unique $m, n$. Composing in both orders gives

$$a = a^{(1-i)m} b^{(1-i)n} = a^{(1-i)m} b^n$$

which by uniqueness gives $n = 0, 1-i = m = \pm 1$.

Subcase: $1-i = m = 1$. Then $ab = ba$ and our group is $\mathbb{Z}^2$.

Subcase: $1-i = m = -1$. Then $b^{-1} ab = a^{-1}$ and our group is the nontrivial semidirect product $\mathbb{Z} \rtimes \mathbb{Z}$ (the fundamental group of the Klein bottle).

Case: $j = \ell = -1$. Then $i - k = k - i = 0$, so $k = i$. This gives $aba^{-1} = a^i b^{-1} = a^{-1} ba$, hence $a^2 b = b a^2$. I suspect that this group is a familiar one but I don't currently recognize it.

Edit: This problem appears to be solved in P. M. Cohn's A Remark on the General Product of two Infinite Cyclic Groups.

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I can’t understand why somebody downvoted this perfectly sane and informative answer. As I always say, some people upvote or downvote at random here. –  Ewan Delanoy Sep 25 '12 at 18:32
    
Case 2 is easily handled: either $k$ is even, so that $a$ generates a normal subgroup, or $k$ is odd, so that $b$ generates a normal subgroup. Both of these are easier to see after quotienting out by $a^2$. –  user641 Sep 28 '12 at 12:32
    
@Steve: I'm afraid I don't see either of these at the moment (but it's quite early). Could you elaborate? –  Qiaochu Yuan Sep 28 '12 at 17:03
up vote 1 down vote accepted

Cohn’s article linked by Qiaochu indeed does provide a complete answer (when you combine it with another paper cited in Cohn’s article's references, namely “Zur Theorie der faktorisierbaren gruppen” by L. Redei in Acta. Math. Acad. Sci. Hung. 1, pp.74-98, 1950 ).

The main result (the “Satz 2” in Redei’s paper) is as follows :

If $G$ is a (noncommutative) group and there are pairs $(a,b) \in G^2$ such that any element of $G$ can be uniquely written in the form $a^{i}b^{j}$ with $i,j \in {\mathbb Z}$, then we can always find one such pair that either

$$ (1) ba=a^{-1}b^{-1}, \ ba^{-1}=ab^{-1} \ (\text{which implies that } \ b^{j}a^{i}=a^{(-1)^ji}b^{(-1)^ij} \ \text{ for any } \ i,j\in{\mathbb Z}) $$

or else, there is an integer $g$ such that

$$ (2) ba=a^{-1}b^{-1}, \ ba^{-1}=a^{-1}b^{1-2g} \ (\text{which implies that } \ b^{j}a^{i}=a^{(-1)^ji}b^{j+gi(1-(-1)^j)} \ \text{ for any } \ i,j\in{\mathbb Z}) $$

If we note denote by $D(G)$ the derived subgroup of $G$ and $p$ the quotient map $G \to \frac{G}{D(G)}$, in case (1) $\frac{G}{D(G)}$ is isomorphic to $\frac{\mathbb Z}{4\mathbb Z} \times \frac{\mathbb Z}{2\mathbb Z}$ (indeed $(i,j) \mapsto ip(a)+j(p(b)-2p(a))$ is one such isomorphism) , and in case (2) $\frac{G}{D(G)}$ is isomorphic to $\frac{\mathbb Z}{(2g-1)\mathbb Z} \times \frac{\mathbb Z}{2\mathbb Z}$ (again, $(i,j) \mapsto ip(a)+j(p(b)-2p(a))$ defines one such isomorphism).

So all those groups are non-isomorphic. Finally, as Redei explains, the semi-direct product is simply the $g=0$ case in this family of groups.

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