# "Are all the words in the free group, or just th..."

https://arbital.com/p/5jh

by Eric Rogstad Jul 22 2016

We can use the freely reduced words to construct the free group on a given set $X$; this group has as its elements the words over $X \\cup X^{-1}$, and as its group operation "concatenation followed by free reduction" $$that is, removal of pairs $r r^{-1}$ and $r^{-1} r$$$\. The notion of "freely reduced" basically tells us that $r r^{-1}$ is the identity for every letter $r \\in X$, as is $r^{-1} r$; this cancellation of inverses is a property we very much want out of a group\.

Are all the words in the free group, or just the freely-reduced words?

If the latter, does saying that the group operation involves free reduction imply that only freely-reduced words will end up in the free group? Because, by default I would assume that "the group has as its element the words over $X \cup X^{-1}$" means that all the words (including non-freely-reduced ones) are in the free group.

Patrick Stevens

You're quite right to flag this up; I was being sloppy. There are three main ways to construct the free group, and I've kind of mixed together the two of them which are most intuitive. I'm trying not to simply define the free group here, but you're right that I've done it confusingly. I'll fix it.

Patrick Stevens

It sounds like you didn't already know what the free group is; in that case (and even if you did already know), it's very gratifying to know that someone is actually reading this carefully!

That's right. I know very little group theory.

I was just remarking to Stephanie how I was able to understand everything on this page, right before I got to that sentence that I found confusing :-)

Patrick Stevens

Are you otherwise broadly Math 3? It would be good to have a guinea pig for group theory.