Is there a Hausdorff topological group $(G,\cal T)$ such that $G$ is a solvable group with a cardinality strictly greater than $\frak c$ and such that there is not any nontrivial (not necessarily Hausdorff) group topology $\cal S$ on $G$ with $\cal S\subsetneqq T$?


There not such topological group if "solvable" is replaced with "abelian". 



An atomic solvable Hausdorff topological group with a cardinality greater than that of real line
**Theorem** (suggested by I.V.Protasov). Every solvable Hausdorff topological group $G$ is topologically solvable in the sense that $G$ contains an increasing sequence of *closed* subgroups $\{1\}=G\_0\subset G\_1\subset\dots\subset G\_n=G$ such that for every $i\le n$ the subgroup $G\_{i-1}$ is normal in $G\_i$ and the quotient group $G\_i/G\_{i-1}$ is abelian.


*Proof:* The group $G$, being solvable, contains an increasing sequence of subgroups $\{1\}=G\_0\subset G\_1\subset\dots\subset G\_n=G$ such that for every $i\le n$ the subgroup $G\_{i-1}$ is normal in $G\_i$ and the quotient group $G\_i/G\_{i-1}$ is abelian.


For every $i\le n$ let $\bar G\_i$ be the closure of the group $G\_i$ in $G$. It can be shown that for every $i\le n$ the normality of the subgroup $G\_{i-1}$ in $G\_i$ implies the normality of the its closure $\bar G\_{i-1}$ in the closure $\bar G\_i$ of $G\_i$. The quotient group $\bar G\_i/\bar G\_{i-1}$ is abelian since is contains a dense abelian group $G\_i/(G\_i\cap\bar G\_{i-1})$ (which is a homomorphic image of the abelian group $G\_i/G\_{i-1}$). So, $G$ is topologically solvable.


**Corollary 1.** Each non-abelian solvable Hausdorff topological group contains a proper non-trivial closed normal subgroup and hence cannot be topologically simple.


The following corollary answer the question of @iranano.


**Corollary 2.** Each solvable Hausdorff topological group $(G,\tau)$ of cardinality $|G|>\mathfrak c$ admits a non-trivial (not necessarily Hausdorff) group topology $\sigma\subsetneq \tau$.


*Proof:* If $G$ is not abelian, then by Corollary 1, $G$ contains a proper non-trivial normal subgroup $H$ and then $\sigma:=\{UH:U\in\tau\}$ is the required weak (non-Hausdorff) group topology on $G$.


So, we assume that $G$ is abelian. If $G$ is not precompact, then by the famous Prodanov-Stoyanov Theorem, $G$ is not minimal and hence admits a strictly weaker Hausdorff group topology $\sigma\subsetneq \tau$.


It remains to consider the case of a precompact abelian group $G$. In this case the completion $\bar G$ of $G$ by the two-sided uniformity is compact, so characters separate points of $\bar G$. Consequently, there exists a non-trivial continuous homomorphism $\chi:G\to\mathbb T=\{z\in \mathbb C:|z|=1\}$ into the circle group. Since $|G|>\mathfrak c=|\mathbb T|$ the kernel $H=\chi^{-1}(1)$ is a non-trivial closed proper subgroup of $G$. Then $\sigma:=\{UH:U\in\tau\}$ is the required weak (non-Hausdorff) group topology on $G$.


