What is this "Schreier factor function"? I'm now reading the paper "Automorphisms of Group Extensions" by C. Wells [Trans. Amer. Math. Soc. 1971]. There is a paragraph mentions about the Schreier factor function $\mu\colon\Pi^2\to G$ ($\Pi$ and $G$ are groups) satisfying the following relations:
$\mu(x,yz)+\mu(y,z)=\mu(xy,z)+\mu(x,y)z$ where $x,y,z\in\Pi$, and
$(ax)y=[a(xy)]^{\mu(x,y)}=-\mu(x,y)+a(xy)+\mu(x,y)$ where $a\in G$, $x,y\in\Pi$. 
I tried to find more properties of this function but I found nothing.
Could anyone provide me the full definition and properties of this fuction?
 A: You are missing out some key information! Specifically, you are missing out the fact that there is the following short exact sequence. $$1\rightarrow G\rightarrow E\rightarrow\Pi\rightarrow 1$$
($(G, E, \Pi)$ is the notation Wells uses - I would lean towards $(N, G, Q)$, but anyway...). You are also missing certain bits of notation. I will explain the notation below when I come to them. Now, I am unsure of any additional properties (I have written a short bit on the properties you have), but I can tell you the definition. Indeed, it is in Wells' paper, but is slightly disguised (that is, it doesn't say "Definiton:...").
Fix a transversal $T$ for $\Pi=E/G$, so $T:\Pi\rightarrow E$. Now, Wells writes $+$ for multiplication in $G$ and $E$ but I shall use $\times$ (because generally $+$ implies commutative, but that is not the case here). I will use $\cdot$ for multiplication in $\Pi$.
Now, if $h\in E$ then it can be written as $h=T(x)\times a$ for some $x\in \Pi$ and some $a\in G$, simply because $G$ is normal. Suppose $h, k\in E$, then $h\times k$ has this form also, so
$$\begin{align*}
h\times k
&=(T(x)\times a)\times (T(y)\times b)\\
&=T(z)\times c
\end{align*}$$
So, the question is "What are $z$ and $c$?" Well, $z=xy$ (again, because $G$ is normal). However, from the information we have we cannot determine $c$ (in general). This is where the Schreier Factor Function comes in. It is precisely the function such that
$$\begin{align*}
(T(x)\times a)\times (T(y)\times b)
&=T(x)\times T(y)\times T(y)^{-1}\times a\times T(y)\times b\\
&=T(x\cdot y)\times\mu(x, y)\times a^{T(y)}\times b
\end{align*}$$
Note that $a^{T(y)}:=T(y)^{-1}\times a\times T(y)\in G$ as, once again, $G$ is normal in $E$.

In summary, $\mu(x, y)$ is the function $\mu: \Pi\times\Pi\rightarrow G$ such that $$T(x)\times T(y)=T(x\cdot y)\times \mu(x, y)$$ in $E$, where $T$ is a transversal function for $\Pi\cong E/G$.

Note that the transversal function $T$ matters, that is, $\mu$ is dependent on $T$. For example, take $E=\mathbb{Z}_6=\{0, 1, 2, 3, 4, 5, 6\}$ and take $G=\langle 2\rangle$, then $\Pi=\{0+G, 1+G\}$. We could take the transversal function to be the function $T_a$ such that $T_a(0+G)=0$ and $T_a(1+G)=1$ or to be the function $T_b$ such that $T_b(0+G)=2$ and $T_b(1+G)=1$ (note that there are many more choices!). Then
$$\begin{align*}
1=0+1&=T_a(0+G)+T_a(1+G)\\
&=T_a(1+G)+\mu_a(0+G, 1+G)\\
&=1+\mu_a(0+G, 1+G)\\
\end{align*}$$
and so $\mu_a(0+G, 1+G)=0$. On the other hand,
$$\begin{align*}
3=2+1&=T_b(0+G)+T_b(1+G)\\
&=T_b(1+G)+\mu_b(0+G, 1+G)\\
&=1+\mu_b(0+G, 1+G)
\end{align*}$$
and so $\mu_b(0+G, 1+G)=2$. Therefore, $\mu_a\neq\mu_b$.
Finally, I should say something about the properties you have written. Specifically, concatenation is not multiplication. Rather, when Wells writes $ax$ for $a\in G$, $x\in\Pi$, this is shorthand for $T(x)^{-1}\times a\times T(x)$, so when you write $\mu(x, y)z$ this is actually $T(z)^{-1}\times\mu(x, y)\times T(z)\in G$, and so on.
