Reduced free group

Question

Let $G$ be a $d$-generated group. Then my first question is how to see free reduced group $FV(G)$ in the variety containing $G$. What I understood is: "Let $W \subset F_d$ ($d$-generated free group) be the collection of words which are laws for $G$, i.e. for every $w \in W$ image of word map $w$ on $G$ is identity. Let $F_m$ be m-generated free group and $S$ be the union of images of words $w \in W$ in $F_m$. If $FV(G)$ is the $m$-generated reduced free group then $FV(G) = \frac{F_m}{\langle S \rangle}$."

My second question is when $G$ is finite group then is it true that finitely generated free reduced group $FV(G)$ will also be finite? Where can I read more about free reduced group?

Thank you!

Answer 1

I'll give a more detailed answer. My comment above about your first question was wrong because I read it too quickly and didn't catch the difference between $m$ and $d$. Your description of the free object in question 1 is off.

Take the cyclic group $C_n$ of order $n\geq 3$. Then $C_n$ is $1$-generated and the laws it satisfies in one variable are all consequences of $x^n=1$. But it also satisfies $xy=yx$, which is not a consequence of this law when $n\geq 3$.

If $A$ is a universal algebra, then the variety generated by $A$ is defined by all laws satisfied by $A$. So you should factor the free group out by the verbal subgroup generated by all laws satisfied by $G$.

For the second question, the answer is yes as I said in the comment. If $A$ is a universal algebra, then the free object in the variety generated by $A$ on a generating set $X$ embeds in $A^{A^X}$ via the map $\psi\colon X\to A^{A^X}$ with $\psi(x)(f)=f(x)$. This $X$ generates a relatively free object because any mapping $f\colon X\to A$ factors through $\psi$ by projecting the the $f$-coordinate and so every map from the free object on $X$ in the signature of $A$ to $A$ factors through the map induced by $\psi$. Now it is easy to verify that $\psi(X)$ generates a relatively free algebra in the variety generated by $A$ using the description as quotients of subalgebras of direct products of copies of $A$.

Answer 2

The relative free group of rank $n$ corresponds directly to a set of maps from $G^n\rightarrow G$. These maps are the coset representatives of $F_n/K$ where $K$ is the group of $n$-variable laws of $G$.

If $G$ is finite, then there are only finitely many such maps and hence the relative free group itself is finite.