Free groups quickfur Junior Member Posts: 32 Threads: 9 Joined: Dec 2006 Reputation: 0 2007-12-07, 16:12 Hi, just a quick question about free groups. I'm studying a class of groups which resemble free groups with a set C of generators (possibly infinite), except that every generator is its own inverse. (But non-generators, i.e., elements not in C, may not necessarily have this property.) Is there a name for this kind of "almost-free" group? Is it isomorphic to actual free groups by any chance? Are there well-known properties of such groups? Thanks. bo198214 Administrator Posts: 53 Threads: 7 Joined: Dec 2006 Reputation: 0 2008-01-19, 14:26 quickfur Wrote:Hi, just a quick question about free groups. I'm studying a class of groups which resemble free groups with a set C of generators (possibly infinite), except that every generator is its own inverse. (But non-generators, i.e., elements not in C, may not necessarily have this property.) Is there a name for this kind of "almost-free" group? Is it isomorphic to actual free groups by any chance? Are there well-known properties of such groups? Thanks. If every generator is its own inverse than the inverse of a word is the reversed word, $(abc)^{-1}=c^{-1}b^{-1}a^{-1}=cba$. So if you take the free group with the relations $gg=1$ for each generator $g$ then the elements are all words (without inverse letters) with non-repeating letters and with the reversion as inversion. Multiplication is concatenation with removing each $gg$ afterwards. What is your specific problem? quickfur Junior Member Posts: 32 Threads: 9 Joined: Dec 2006 Reputation: 0 2008-01-28, 19:44 bo198214 Wrote:If every generator is its own inverse than the inverse of a word is the reversed word, $(abc)^{-1}=c^{-1}b^{-1}a^{-1}=cba$. So if you take the free group with the relations $gg=1$ for each generator $g$ then the elements are all words (without inverse letters) with non-repeating letters and with the reversion as inversion. Multiplication is concatenation with removing each $gg$ afterwards.Yes, I got this far, and also that the group can be represented by an infinite tree (which I guess would correspond to its Cayley graph) where every node is equivalent under multiplication. The elements of the group can then be considered as "paths" from some arbitrary fixed node (which we may consider to be the "root" node) to some other node. There is always a unique path between any two nodes. Now, given a set of paths relative to a root node R1, we can rewrite the paths to be relative to some other root node R2 by left-multiplying each path with the inverse of R2. This gives us a vector-like algebra of paths, where given a path from A to B and a path from B to C, multiplication yields a path from A to C (which is always unique, since the definition of multiplication removes redundant paths that traverse a set of nodes more than once). We can relocate the origin freely using the left-multiplication-by-inverse technique. Quote:What is your specific problem? Not really a specific problem, just exploring a possible generalization of the natural numbers that has non-unique negatives (e.g., consider the group with 3 generators with the self-inverse property---the group with 2 generators is isomorphic to the integers). I was just wondering if this type of groups have been studied before, so that I don't have to derive their properties myself. I'm interested in generalizing these types of groups to dense (perhaps continuous?) sets that preserves the "pseudo-vector" properties listed above. bo198214 Administrator Posts: 53 Threads: 7 Joined: Dec 2006 Reputation: 0 2008-02-07, 17:34 quickfur Wrote:Not really a specific problem, just exploring a possible generalization of the natural numbers that has non-unique negatives (e.g., consider the group with 3 generators with the self-inverse property---the group with 2 generators is isomorphic to the integers). I was just wondering if this type of groups have been studied before, so that I don't have to derive their properties myself. The group with two generators $a$ and $b$ contains the elements $e,a,ab,aba,abab,\dots,b,ba,baba,\dots$. How is this group isomorphic to the integers (under addition?)? $n=\psi(a)=\psi(a^{-1})=-n$ then $n=0$ and $\psi(a)=0=\psi(b)$ and then $\psi$ is not bijective. quickfur Junior Member Posts: 32 Threads: 9 Joined: Dec 2006 Reputation: 0 2008-02-21, 06:05 bo198214 Wrote:[...] The group with two generators $a$ and $b$ contains the elements $e,a,ab,aba,abab,\dots,b,ba,baba,\dots$. How is this group isomorphic to the integers (under addition?)? $n=\psi(a)=\psi(a^{-1})=-n$ then $n=0$ and $\psi(a)=0=\psi(b)$ and then $\psi$ is not bijective.Hmm you're right. The group operation doesn't correspond with integer addition. But you can create a 1-to-1 map between group members and integers: the only valid strings in this group are alternating a's and b's, so you can designate all strings that begin with a as "positive", and all strings that begin with b as "negative", and then map them to positive and negative integers, respectively, based on their length. (Since there are only two generators and they are idempotent, the string length uniquely determines the string once you've fixed the first element.) « Next Oldest | Next Newest »