In section 7.5 of John Stillwell's 'Naive Lie Theory' he constructs a proof of what he calls the 'Tangent space visibility' theorem: that for any path-connected matrix Lie group $G$ with a discrete center and non-discrete normal subgroup $H$, the tangent space of $H$ at the identity $T_{1}(G)$ is non-trivial. He proves this by using a topology argument that a neighborhood of the identity in $G$ generates $G$ if $G$ is path-connected, which implies that you can find a neighborhood of the identity which contains no element of the center (other than 1) some $B\neq 1$ in $H$ and some $A$ which $B$ does not commute with. Once he has this $A$ he is able to construct a path in $H$ with a non-zero tangent vector, proving the theorem.
I apologize if my explanation of the proof is not sufficient, but perhaps someone is familiar with it or has a copy of the text that they can find it in. Anyway, in problem 7.5.4 we are asked to modify this proof such that $G$ does not have to be path-connected: my understanding of this change is that now we cannot use the '$N_{\delta}(1)$ generates $G$' argument to show the existence of the $A$ which doesn't commute with $B$, we need to find an alternate way to prove that such an $A$ exists. It is here that I am stuck, as I am having trouble reaching a contradiction upon supposing that $B$ commutes with each element of the neighborhood $N_{\delta}(1)$. Any pointers in the right direction would be greatly appreciated.
I think the nondiscrete normal subgroup $H$ must be in the identity component of the group $G$. Because a normal subgroup must include the identity $1$, and for a nondiscrete subgroup, there must be a path between any two elements.
So even if the group $G$ is not path connected, we can also find the neighbour $N_\delta(1)$ in the proof in the identity component of $G$. See the section 8.6 or exercise of section 3.2 for the definition of the identity component.