Prove that a free group of rank $\ge2$ is centerless and torsion-free This exercise is from Rotman's Introduction to the Theory of Groups. It's just as the title states: prove a free group of rank $\ge2$ is centerless torsion-free.
Here, the definition of a free group with basis $X$ is a group $F$ such that given any group $G$ and function $f:X\to G$, there is a unique $\varphi:F\to G$ extending $f$.
The conclusion I'm supposed to make makes perfect sense when we consider what free groups really look like: they are just groups defined on a set of generators with no relations. However, I need to use this specific definition of free group for this problem and I'm having trouble. I've already proved it's centerless but I need some help on torsion-free.
So here's the kind of thing I'm trying: Suppose we have a torsion element, call it $z$. Then $z^n=1$ for some $n$. Define a map $f:X\to F$ by $f(x)=x^n$. Because $F$ is free with basis $X$, this extends to a unique $\varphi:F\to F$. I want to somehow come up with a different map $\psi$ which also extends $f$, contradicting the uniqueness of $\varphi$. I'm not sure how to do this though. Does anybody have any suggestions, or alternative ideas?
 A: First of all, as I said in my comment, you should prove this using the description of $F_X$ as the set of reduced words in the alphabet $X \cup X^{-1}$, as surely is intended. However, instead of this, let me give a geometric proof that $F_X$ is torsion-free, intended for those who know some basic algebraic topology. This solution is a geometric version of what you will be doing combinatorially. 
First of all, let $Y$ denote the bouquet of circles labelled by the elements of $X$ (if $X$ has cardinality $k$, we will have the bouquet of $k$ circles). According to the Van Kampen theorem, $G=\pi_1(Y)\cong F_X$. According to the fundamental theorem of the theory of covering spaces, $G$ acts on the universal cover $T$ of $Y$ as the group of covering transformations of the universal cover $T\to Y$. In particular, no nontrivial element $g\in G$ has a fixed point in $T$. The space $T$ is a graph (since $Y$ is), it is also simply connected, hence, the graph $T$ is a tree (is connected and contains no circuits). I will equip $T$ with the graph metric: The distance between two vertices is the minimal edge-length of the edge-paths connecting these vertices.  Now, given $g\in G$, consider the following function on the vertex set $V$ of $T$:
$$
d(v):= d(v, g(v)). 
$$
Since $d$ is integer-valued, this function attains its minimum at some vertex $v\in V$. Since $g\ne 1$ has no fixed vertices, $d_g>0$ for all $g\ne 1$. Consider now the biinfinite sequence of vertices 
$$
... g^{-2}v, g^{-1}v, v, gv, g^2v, ...
$$ 
Connect the consecutive vertices $v_i, v_{i+1}, i\in {\mathbb Z}$, in this sequence by the shortest edge-paths $v_i v_{i+1}$ in $T$. Let $A$ denote the biinfinite path in $T$ formed by concatenating these edge-paths. By the construction, $g$ preserves $A$. Since $v$ minimizes the function $d$, the path $A$ does not backtrack:
$$
v_{i-1}v_i \cap v_iv_{i+1}= v_i, i\in {\mathbb Z}.
$$ 
Since $T$ is a tree, the map
$$
f: {\mathbb R}\to A, f([i, i+1])= v_i v_{i+1}, i\in {\mathbb Z}, 
$$
is a homeomorphism. By the construction, $g$ acts on the path $A$ via translations:
$$
v_i \mapsto v_{i+1}, i\in {\mathbb Z}. 
$$ 
If $g$ were to have finite order $n$, we would therefore conclude that 
$$
v_n= v_0.
$$ 
But this contradicts the property that $f^{-1}(v_i)< f^{-1}(v_j)$ whenever $i<j$. qed 
The proof that $F_X$ is centerless (if $X$ has cardinality $\ge 2$) is similar: For a nontrivial element $g$ consider the line $A=A_g\subset T$ as constructed above. Then show that 
$$
A_{hgh^{-1}}= h A_g. 
$$ 
Then for each nontrivial central element $g\in G$, the line $A_g$ is preserved by $G$, from which you conclude that $Y\cong S^1$, i.e. $|X|=1$.   
