# Proving things about the free group from the categorical definition.

I'm undertaking A Course in the Theory of Groups by Robinson and I'm looking for some guidance on some of the exercises. Specifically, I'm trying to show that a free group of rank 2 or higher has a trivial center. This is completely obvious in view of the free group as the group of reduced words, but I'm trying to show it using categorical language. The book also asks that I show that the image of the generating set, that is $\operatorname{im} \sigma \subset F$ for $\sigma \colon S \to F$, generates $F$. I believe this is done by using the assumption that the homomorphism $\beta \colon F \to G$ is unique for a given $\alpha \colon S \to G$. It specifically asks that this is shown using the definition.

Lastly, I'm a little confused about the notation in this book. Its writes the composition of functions from left to right, and a function evaluated at a specific element is written with the element on the left with no parenthesis. Is this standard notation in Category theory?

• The reverse composition of notation was popular among some algebraists for a time, perhaps from the fifties to the seventies. It's out of fashion now. The motivation is that we read English from left to right, so it makes sense to start with the argument and then do more things the further right we go. It also makes sense if we write a string of arrows (from left to right) to represent functions $A \overset{f}{\to} B \overset{g}{\to} C$; then one composes this as $fg$ by multiplying the labels (in the same order). – jdc Apr 25 '14 at 4:25
• In the second question, I assume $F$ is a free group and $S$ its free generating set? To help answer the question, what does it mean to say $\mathrm{im}\ \sigma$ generates $F$? – jdc Apr 25 '14 at 4:41
• Yes F is the free group, S is the generating set, and $\sigma$ is the map from S to F. I'm trying to show that $im \ \sigma$, which is a subset of F, generates F in the usual group theoretic sense, that is F is the smallest subgroup that contains $im \ \sigma$. – Ryan Keathley Apr 25 '14 at 4:50

Let $F$ be a free group on $S$ and $i : S \to |F|$ the unit of the adjunction, i.e. what is usally called the inclusion of the basis (here $|F|$ denotes the underlying set of the group $F$). Let $F'$ be the subgroup of $F$ generated by (the image of) $i$. Then we have a factorization $i = |f|k$ for some $k : S \to |F'|$ and a monomorphism $f : F' \to F$. By the universal property of $(F,i)$ there is a homomorphism $g : F \to F'$ such that $|g|i=k$. It follows $|fg|i = |f|k=i=|\mathrm{id}|i$, hence (by uniqueness in the universal property of $(F,i)$) $fg=\mathrm{id}$. Then $f$ is surjective, i.e. $f$ is an isomorphism, i.e. $i$ generates $F$.
I doubt that that there is a purely categorical proof that $Z(F)=1$ when $|S| \geq 2$. For this one needs reduced words (which, by the way, can be deduced from the universal property, as is explained by Serre in his book Trees).