The charge given by a commutator I saw in the text that
$$[Q,X]=cX$$
and says the operator $X$ has charge $c$ under the generator $Q$.
I tried to understand why the coefficient $c$ means the charge. So I used this relation to get the result(using the property of $e^{a}=\lim_{n\rightarrow\infty}(1+\frac{a}{n})^{n}$)
$$e^{iQ} X e^{-iQ}=e^{ic}X$$
which seems like the commutator really shows the charge of the operator $X$ as $c$. Is it right way to understand the statement?
 A: If $Q|q\rangle = q|q\rangle$, then
$$
Q(X|q\rangle)= (QX-XQ+XQ)|q\rangle= ([Q,X] +Xq)|q\rangle=(c+q)X|q\rangle.
$$
So applying $X$ to an eigenstate $|q\rangle$ of $Q$ increases the eigenvalue, or "charge," by $c$. We can  say that $X$ is giving a charge $c$ to $|q\rangle$.
A: The basic idea is that we transform in proportion to such commutators to produce a symmetry. It may help to discuss an example that's a little more complicated than what you've described.
The infinitesimal BRST transformation $X\mapsto X+\delta X$ is a symmetry of the BRST-quantised Yang-Mills Lagrangian. The BRST operator $Q$ (also called the BRST charge) satisfies $\delta b=\theta [Q,\,b],\,\delta f=\theta\{ Q,\,f\}$ for an infinitesimal constant Grassmann number $\theta$, with arbitrary bosonic field $b$ and fermionic field $f$. The need for anticommutators with fermions comes from $Q$ being fermionic (which is why $\theta$ needs to be a Grassmann number to make $\delta$ bosonic); if $Q$ were bosonic, in which case we wouldn't be discussing the BRST example, it'd be commutators all the way.
We can verify $\delta(XY)=(\delta X)Y+X\delta Y$ for fields $X,\,Y$ of definite exchange symmetry, and $\delta$ commutes with $\partial_\mu$.
