I'm new to profinite groups and wish to prove the following, which is a claim I found in many sources, but it is probably so simple that nobody even proves it. Let $G$ be a group and $\hat{G}$ be its profinite completion. We have a canonical inclusion $G\to\hat{G}$ and the image of $G$ in $\hat{G}$ is dense.
Why is the image of $G$ dense in $\hat{G}$? Any hints?
If $N$ is a normal subgroup of $G$ of finite index then one has a morphism $f_N\colon\hat G\rightarrow G/N$ by definition of the profinite completion of $G$. Let $\hat N$ be the kernel of $f_N$. The system of subsets $\hat N$ is a basis of open neigborhoods of the neutral element of $\hat G$. A basis for the topology of $\hat G$ is the system of all translates $\hat x\hat N$, where $\hat x\in\hat G$ and $N$ a normal subgroup of $G$ of finite index. We have to show that each subset $\hat x\hat N$ of $\hat G$ intersects $G$. Let $x$ be an element of $G$ such that $x=f_N(\hat x)$ in $G/N$. Since $$ f_N^{-1}(f_N(\hat x))=\hat x\hat N, $$ one deduces that $x\in\hat x \hat N$. This proves that $$ \hat x\hat N\cap G\neq\emptyset $$ for all $\hat x\in \hat G$ and all normal subgroups $N$ of $G$ of finite index.