Free groups and derivative Per definition a derivative on a group $G$ is a mapping $D:G\rightarrow\mathbb{Z}G$ such that $D(gh)=D(g)+gD(h)$. Now my question:
uppose $G$ is a free group $F=F(X)$ with $X$ a finite set and suppose $D$ is any derivative. Why is $D$ completely determined by the values $D(x)$ (for $x\in X$) that it takes on the generators. Why holds the formula:$$D(w)=\sum_{x\in X}{a_x(w)D(x)},\ \ \ a_x(w)\in\mathbb{Z}F$$ and what are the coefficients? Here $\mathbb{Z}F$ is the groupring of $F$. Can someone help me with this question?
For more about a derivative defined above:


*

*$D(g^{-1})=-g^{-1}D(g)$ 

*$D(g^n)=\frac{g^n-1}{g-1}D(g)$

 A: Because, $F$ is free on the set $X$, there are no relations among words (beyond those that make it a group:  associativity, identity, inverses).  You can prove your formula by induction on the length of the word $w$.
First, can you show that $D(1) = 0$, where $1 \in F$ is the identity?  Hint: $1 = 1 \cdot 1$.
While you're at it, why not observe that a word of length $1$ is just $w = x$ or $w = x^{-1}$ for $x \in F$.  Do you see why $D(w)$ is of the correct form?
Now, for the inductive step, any word can be written
$$
w = w_1 \cdots w_{n - 1} \cdot w_n = (w_1 \cdots w_{n - 1}) \cdot w_n.
$$
Use the Leibniz rule and the inductive hypothesis, and you've got your result.  Can you see how to finish it?
A: The reason $D$ is defined on values from $X$ is because every element of $F$ is some word in the letters from $X$ and you can always expand any word using the rules of the derivative.  For example:
$$D(ab^{-1}c) = D(a) + aD(b^{-1}c)$$
$$= D(a) + a(D(b^{-1}) + b^{-1}D(c))$$
$$= D(a) + aD(b^{-1}) + ab^{-1}D(c)$$
$$= D(a) + a(-b^{-1}D(b)) + ab^{-1}D(c)$$
$$= D(a) - ab^{-1}D(b) + ab^{-1}D(c)$$
I don't know of a formula that just gives you the coefficients (there probably is one but there's no guarantee that it's nice).  Instead you can find the coefficients by continually expanding terms until the only arguments to the function $D$ are single letters from $X$.  So, for example, in the above we found that:
$$a_a(ab^{-1}c) = 1$$
$$a_b(ab^{-1}c) = -ab^{-1}$$
$$a_c(ab^{-1}c) = ab^{-1}$$
