I read Atiyah Macdonald Introduction to commutative algebra. In this book at p.105, the author says
If $\phi : G \rightarrow \hat{G}$ is an isomorphism we shall say that $G$ is a complete.
I think this $\phi$ is defined as $x \mapsto [\{x\}_{n=1}^{\infty}]$. $[\{x\}_{n=1}^{\infty}]$ is an element in $\hat{G}$ and expresses an equivalence class of cauchy sequences which is equivalent to the constant sequence $\{x\}_{n=1}^{\infty}$. In this book, a cauchy sequence and its equivalent class in topological group is defined as the following.
A Cauchy sequence in $G$ is defined to be a sequence $(x_{\nu})$ of elements of $G$ such that , for any neighborhood $U$ of $0$, there exists an integer $s(U)$ with the property that $x_{\mu}-x_{\nu} \in U$ for all $\mu , \nu \geq s(U)$.
Two Cauchy sequences are equivalent if $x_{\mu} - x_{\nu} \rightarrow 0$ in $G$.
The author says, in p.105,
Thus (10.5) asserts that the completion of $G$ is complete.
However, we can just say $\hat{\hat{G}}$ is isomorphic (as a topological group) to $\hat{G}$ in (10.5). I do not know $\phi : \hat{G} \rightarrow \hat{\hat{G}}$ is an isomophism.
And, in (10.5), I think this book may use the following fact.
If two topological spaces $G$, $H$ are isomorphic each other, $\hat{G}$ and $\hat{H}$ are also isomorphic each other.
This may be true in topological group. But in general topology, I know this is NOT true.For example $(0,1)$ and $\mathbb{R}$. That is because they are isomorphic but the completion of the former is $[0,1]$, the latter is $\mathbb{R}$. They are not isomorphic as topology.
Therefore if there exists two topological group $G$ and $H$ such that $G \cong H$ (as topological group) and $G$ is complete but $H$ is not, we can get $\hat{H} \cong \hat{G} \cong G \cong H$ as topological group. So the $H$ is a topological group which is not complete but isomorphic to its completion $\hat{H}$.