Lie algebra of a horizontal vector field and a fundamental vector field Let $G$ be a Lie group, $\mathfrak{g}$ be its Lie algebra, $(P,M,G)$ be a principal $G$ bundle with connection $\omega$.
By a horizontal vector field on $P$, we mean a vector field $X:P\rightarrow TP$ such that $X(p)\in T_pP$ is actually a horizontal vector for each $p\in P$.
Given an element $A\in \mathfrak{g}$ we define a vector field on $P$, calling it fundamental vector field associated to $A$, as $X(p)=(R_p)_{*,e}(A)$ where $R_p:G\rightarrow P$ is defined as $g\mapsto pg$. Just to emphasise it’s relation with $A$ we denote it by $A^*$.
Now the question is to prove that if $X$ is a horizontal vector field on $P$ and $A^*$ is a fundamental vector field associated to $A$, then the Lie brakcte$[X,A^*]$ is a horizontal vector field.
I do not really understand how can one check something is a horizontal vector field. 
There is a proof which uses the notion of lie derivative relation with lie bracket. I am getting confused with that. 
Are there any other methods to prove this..?
 A: I am not sure whether there is a direct/alternative proof, but as far as I can see, the question is not really related to the fact that you deal with horizontal fields. What is actually going on here is that you are dealing with a $G$-invariant distribution $E\subset TG$. This meanst that denoting by $r^g:P\to P$ the principal right action of $g\in G$ and by $E_u\subset T_uP$ the subspace given by the distribution in a point $u\in P$, then you get $E_{u\cdot g}=T_ur^g(E_u)$. (The fact that the horizontal distribution is $G$-invariant is exactly what distinguishes a principal connection from a general fiber bundle connection on a principal bundle.)
Now there is an infinitesimal version of $G$-invariance of a distribution $E$. This is exaclty that for each section $\xi$ of $E$ and each fundamental vector field $A^*$, also $[A^*,\xi]$ is a section of $E$. The point about this is indeed that the flow of $A^*$ up to time $t$ is given by $r^{\exp(tA)}$. This shows that pulling back $\xi$ by this flow, you always get a section of $E$ and the relation between this pullback and the Lie bracket implies directly that $[A^*,\xi]$ has values in $E$. 
Alternatively, you can deduce the statement from properties of $G$-invariant sections. It is easy to see that any $G$-invariant distribution admits local frames consisting of $G$-invariant sections. (In your case, one could take the horizontal lifts of the elements of a local frame on the base.) Denoting such a frame by $\eta_i$, you can locally write $\xi=\sum_if_i\eta_i$ for smooth functions $f$. Since each $\eta_i$ is invariant its pullback along $r^{\exp(tA)}$ is again $\eta_i$ and this implies that $[A^*,\eta_i]=0$ for all $i$. But then 
$$
[A^*,\xi]=\textstyle\sum_i [A^*,f_i\eta_i]=\sum_i(A^*\cdot f)\eta_i
$$
and this evidently is horizontal. 
