6

If $F$ is a free group then $g^2=h^2$ implies $g=h$ for $h,g\in F$.

I've been trying to prove this given the definition of a free group $F$: given group $F$ and subset $X\subseteq F$, $F$ is free over $X$ if for any group $G$ and function $\theta: X \to G$, there exists a unique homomorphism $\alpha:F\to G$ such that $\alpha(x)=\theta(x)$ for all $x\in X$.

The definition doesn't leave me much to work with: I've attempted to define a function $\theta:X\to F$ and then use the definition to take the given $\alpha$ and somehow arrive at $h=g$ but I've been unsuccessful.

Nir Goldberg
  • 61
  • 1
    For this question it is easier to use the fact that the elements of the group with free generating set $X$ are the reduced words over $X \cup X^{-1}$. That is, strings of letters in $X \cup X^{-1}$ that contain no adjacent $x$ and $x^{-1}$. Then each element has the form $\alpha\beta\alpha^{-1}$, where $\alpha$ is a reduced word and $\beta$ is a cylically reduced word, and the square of this element is $\alpha \beta^2 \alpha^{-1}$. – Derek Holt Jan 18 '15 at 13:32
  • @Derek I'm not sure if I'm allowed to use this - we've learned the definition, but we haven't learned that every free group is of such a form. Can it be proven using the definition outlined in my question, or alternatively, the definition that if $F$ is free over $X={x_1,...}$ then $F={x_{e_1}^{i_1}...x_{e_n}^{i_n}|x_{e_i}\in X, i_j \in N}$? – Nir Goldberg Jan 18 '15 at 13:35
  • @NirGoldberg Yes, you can show that reduced words form a group and that this group has the universal property, hence is canonically isomorphic to any "other" free group over $X$. – Hagen von Eitzen Jan 18 '15 at 13:50
  • The trick with the writing $\alpha \beta \alpha^{-1}$ is that you have the reduced writing of all the powers $\alpha \beta^n \alpha^{-1}$. – orangeskid Jan 18 '15 at 14:41
  • If I write each word as $\alpha\beta\alpha^{-1}$, how does this help me? Eventually I get $\alpha_1\beta_1^2\alpha_1^{-1}=\alpha_2\beta_2^2\alpha_2^{-1}$ which doesn't look like enough to prove $h=g$. – Nir Goldberg Jan 18 '15 at 18:25
  • Just noticed that you are only allowed to use (one) definition of the free group. That is nontrivial. If you are using a homomorphism induced by some map of generators how can you be sure that the images of $g$ and $h$ are different? Also you need to say what groups have you studied before the free group.

    The easiest proof I see is to first prove that free groups are totally orderable. It can be done in many easy ways just by using the definition.

     – markvs Jul 27 '20 at 01:08

2 Answers2

1

You can use the following fact: If $F$ is a free group, $x\in F$ and $x\neq e$, then $C_F(x)$ is infinite cyclic group.

Note that $g^2=h^2=e$ implies $g=h=e$, since there is no non-trivial element of finite order in free group. Hence, assume that $g^2=h^2\neq e$, and let $C_F(g^2)= \langle u\rangle$. Trivially, $g,h\in C_G(g^2)= C_F(h^2)$, hence $g=u^p$ and $h= u^q$. Then $u^{2p}= g^2=h^2= u^{2q}$, so $2p=2q$ since $\langle u\rangle$ is infinite. Therefore, $p=q$ and $g= u^p= u^q= h$.

SMM
  • 4,579
1

This is true for every totally orderable group: if $x\ne y$ then WLOG $x<y$ and $x^2<y^2$.

The fact that free groups are orderable is well known. Total ordering on the free group

markvs
  • 19,896