Let $G$ be a finite group and $\bar\rho:G\rightarrow GL_2(k)$ be a residual representation, where $k$ is a field of characteristic $p$ that is the residue field of a complete discrete valuation ring $R$ of characteristic zero. If the order of $G$ is relatively prime to $p$, then there exists a lift $\rho:G\rightarrow GL_2(R)$ and it is unique up to conjugation by a matrix in $R$ that is identity modulo $m$. Here, $m$ is a unique maximal ideal of $R$ so that $k\cong R/m$.
Author proved this lemma briefly and i have a problem to understand it.
Proof is following: One constructs inductively lifts to $GL_2(R/m^n)$. The obstruction for a lift from the case $n$ to the case $n+1$ is given by an element in $H^2(G,M_2(m^n/m^{n+1}))$. We can check that second cohomology vanishes.(Also, we can check that first cohomology group is zero, and thus any two lifts are conjugate.) Thus one can produce an inverse system of lifts which together provides us with the desired lift into $GL_2(R)$.
My questions are following:
(i): Assume that we have a lift $\rho_n:G\rightarrow GL_2(R/m^n)$. Why second comology group $H^2(G,M_2(m^n/m^{n+1}))=0$ implies the existance of lift $\rho_{n+1}:G\rightarrow GL_2(R/m^{n+1})$? I just know the basic definitions for group cohomology. Likewise i don't understand why first cohomology group is zero implies that any two lifts are conjugate.
(ii): Since $R\cong\lim_{\leftarrow n}R/m^n$ and $(\rho_n(g))_{n}\in\lim_{\leftarrow n}R/m^n$, define $\rho(g):=(\rho_n(g))_{n}$. Is it right??