on question about notation an inner automorphism Let $G$ be group and we define  $x^g=g^{-1}xg$ and $[a,b]=a^{-1}b^{-1}ab$  for $x,g,a,b\in G$.
If $\alpha_{g}$ is an inner automorphism, then do 
$$(x)\alpha_{g}=g^{-1}xg\hspace{5mm} or\hspace{5mm} \alpha_{g}(x)=g^{-1}xg?$$
Please help me.
Thank you
 A: You have the choice.  However this somewhat depends on  how do you compose the maps. 
Having two maps $f : E \rightarrow F$ and $g : F \rightarrow G$.  One can define a third map from $E$ to $G$, some people denote it by $f \circ g$ (or simply $fg$) and the others by $g \circ f$.  Now if $x \in E$, you can denote its image (say) under f, by $f(x)$ or $(x)f$ (or similarly $x^f$). You have all the following possibilities to write the image of $x$ under the third map :
(1) $f \circ g (x) = g(f(x))$
(2) $(x)f \circ g = ((x)f)g$ 
(3) $g \circ f (x) = g(f(x))$
(4) $(x)g \circ f$ = ((x)f)g$
You can see that the notation $f \circ g$ is more appropriate to writing the map on the right (compare (1) and (2)).  Similarly you may notice that (3) is better than (4).
Most of the group theorists (or I think the English school) use the notation $f \circ g$ to denote the composed map, thus it is appropiate in this case to write $(x) \alpha_g$.  On the other hand the second is more appropriate for the French school, moreover they define the inner automorphism induced by $g$ as $\alpha_g(x)=gxg^{-1}$.  I left the reason as a (simple) question to the OP.  
