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 ?
Products of copies of $\mathbb Z$ that are not semi-direct
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0What about the free product $\,\Bbb Z *\Bbb Z\,$, which is nothing else that the free group in two generators...? – 2012-09-21
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0@DonAntonio: can't write everything as $a^ib^j$ in that case – 2012-09-21
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0Yes, I just noticed that: it's just one single element of the form $\,a^ib^j\,$ and not a (finite) product... – 2012-09-21
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0Not all such groups are products of $\mathbb{Z}$, for example the group $\mathbb{Z}[\frac{1}{2}]\rtimes \mathbb{Z}$ where the action is given by multiplication by 2. – 2012-09-21
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2@Mustafa: what are the two elements $a, b$ in this case? – 2012-09-21
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0@Qiaochu There are no such elements. I did not read the question carefully enough. – 2012-09-21
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0How does $aba^{-1}=b^j$ implies the given hypothesis? – 2012-09-22
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0@Loki: write it as $ba^{-1} = a^{-1} b^j$. This relation lets you move any power of $a$ past any power of $b$ (while changing the power of $b$). – 2012-09-22
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2It 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. – 2012-09-22
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0@Derek: of course you're right. I think this observation actually ends up making the problem easier but I haven't worked out the details yet. – 2012-09-25
2 Answers
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.
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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0I can’t understand why somebody downvoted this perfectly sane and informative answer. As I always say, some people upvote or downvote at random here. – 2012-09-25
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0Case 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 $a^2$. – 2012-09-28
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0@Steve: I'm afraid I don't see either of these at the moment (but it's quite early). Could you elaborate? – 2012-09-28