A subgroup, $H\subset G$, is called core-free if $H$ contains no non-trivial normal subgroups of $G$. One can define $\text{Core}_G(H)$ to be the largest normal subgroup of $G$ contained in $H$. The core-free subgroups, $H\subset G$, are exactly those with $\text{Core}_G(H)=1$.
I am interested in characterizing all Lie groups, $G$, such that all of its core-free subgroups are trivial. That is, I am looking for all groups $G$ where $\text{Core}_G(H)=1$ implies $H=1$.
Example 1: Every subgroup, $H\subset G$, of an abelian group, $G$, is also a normal subgroup, $H\lhd G$. Normal subgroups have $\text{Core}_G(H)=H$. Thus, when $G$ is abelian we have that $\text{Core}_G(H)=1$ is logically equivalent to $H=1$.
Example 2: The above proof actually holds for all Dedekind groups, $G$.
Example 3: A non-Dedekind example is discussed here.
But is there a full characterization of these groups? I am particularly interested in the case where $G$ is a Lie group although a result about finite groups could also be interesting.
Best Answer
If $G$ is a positive dimensional compact Lie group and each subgroup $H$ of $G$ with trivial core is trivial, then $G$ must be abelian: otherwise $G$ contains a subgroup $K$ isomorphic to $\mathrm{SU}_2$ or $\mathrm{SO}(3)$. In either case we obtain a subgroup with trivial core which is not trivial: for instance, in $\mathrm{SU}_2$ we consider the matrix $$g=\left( \begin{matrix} \zeta & 0 \\ 0 & \zeta^{-1} \end{matrix} \right)$$ with $\zeta$ a primitive $3$rd root of one. The subgroup $H=\{1,g,g^2 \}$ generated by $g$ has trivial core in $K$ and hence in $G$.