I'm interested in knowing how, given a set $S$, one can modify the free group $F(S)$ by adding one or more commutation rules and get a new group. For instance, adding the commutation rule:
$$\forall x,y\in S, xy=yx$$
yields the free abelian group generated by the base $S$. So I'm wondering: does any commutation rule work? For instance, if I say
$$\forall x,y\in S, xy=yx^2$$
or even that, for a fixed $z\in S$,
$$\forall y\in S,y\neq z, zy=y^{-1}z^2$$
the first is only possible in the trivial group ($x=x1=1x^2=x^2$) but the second does seem to make sense, at least at a first glance, Which, in general, of these rules produces a new group as compared to those which induce a contradiction or work only in the trivial group? Is there for example a big family of such rules that will tune the original free group into a different, non trivial one?
[Note: I'm not sure what a proper definition of "commutation rule" could be. I just threw some random examples that fit my intuition for such a term.]
In particular, take the following example.
As a continuation of my previous post (Extending automorphism to inner automorphism?), I think I've come to a solution to the following problem.
- Let $H$ and $K$ be subgroups of a finite group $G$ and assume $H$ is isomorphic to $K$. Prove that there exists a group $\bar G$ containing $G$ as a subgroup, such that $H$ and $K$ are conjugate in $\bar G$.
(Problem 56 on this problem set.)
Let $\phi:H\to K$ be an isomorphism. My idea is to modify the free group in $G\cup\{x\}$, $F(G\cup\{x\})$, by letting all the elements of $G$ multiply themselves as usual and by adding the following commutation property.
$$\forall h\in H,xh=\phi(h)x$$
Call this new group $\bar G$. This way, $G\leq\bar G$ and for all $h\in H$, $^xh=\phi(h)$ and in particular $^xH=K$.
[Note: I know I've added another rule (the one about the elements of $G$ multiplying themselves as usual) but that one could radically be seen as a "commutation property": $\forall g,h\in G, gh=g\cdot_Gh$.]
As I'm not sure if this definition is enough by itself, here's a more schematic definition of $\bar G$.
$$\bar G=\{g_1x^{k_1}g_2x^{k_2}\ldots g_nx^{k_n}:g_i\in G,g_{i>1}\notin H,k_i\in\mathbb{Z},k_{i<n}\neq 0\}$$
Two elements are equal if they have the same canonical form, i.e. the same expression as an interleaved product of elements of $G$ and powers of $x$, subject of the same conditions that I included in the previous definition. The product of two elements is carried out by concatenating their canonical forms and then commuting all the $g_i$'s that belong to $H$, possibly cancelling out some powers of $x$ in the process: since there are only finitely many powers of $x$ in their expressions, this process necessarily ends, yielding the canonical form of their product.
- Does my "commutation rule" actually generate a new group here?
- If so, is it a special case or is there a bigger family of "commutation rules" that will produce a group?
Best Answer
What you describe is, in essence, a group presentation. The study of presentations is known as combinatorial group theory.
The theory can get technical very quickly, but the idea is that you take a free group $F=\langle X\rangle$ generated by a set $X$ and a set $R$ of elements of $F$ written in terms of $X$, then the group you're after is
$$G\cong F/\langle\langle R\rangle \rangle,$$
where $\langle\langle R\rangle \rangle$ is the normal subgroup of $F$ generated by $R$. If you want $u=v$, then put $r=uv^{-1}\in R$.
The notation is
$$G\cong\langle X\mid R\rangle,$$
with set notation omitted if the context is clear. For example,
$$\Bbb Z_2\times\Bbb Z_3\cong \langle a,b\mid a^2, b^3, aba^{-1}b^{-1}\rangle,$$
also written
$$\Bbb Z_2\times\Bbb Z_3\cong \langle a,b\mid a^2, b^3, ab=ba\rangle.$$
Every group has a presentation; in fact, each group has infinitely many of them!
Just let $$R\supseteq \{\varphi (a,b,c,\dots)(\psi(x,y,z,\dots))^{-1}\mid a,b,c,\dots, x,y,z,\dots\in F\}$$ to get $$\varphi (a,b,c,\dots)=\psi(x,y,z,\dots)$$ for all elements in the group. Remember: it works for any subset $R$ of $F$.