Is norm-continuous representation factored through a Lie quotient group?

Question

I asked this 11 days ago at MSE, but there was no answer, I hope people here could help.

Let $G$ be a locally compact group, and $X$ a Hilbert space. A unitary representation $\varphi:G\to B(X)$ is said to be norm-continuous, if it is continuous with respect to the norm in $B(X)$: $$ x_i\to x\quad\Longrightarrow\quad \lVert\varphi(x_i)-\varphi(x)\rVert\to 0. $$ My question is the following:

Is it true that each norm-continuous unitary representaion $\varphi:G\to B(X)$ can be factored through a Lie quotient group?

I.e. is there a Lie quotient group $G/H$ such that $$ \varphi=\varphi_H\circ\pi_H, $$ where $\pi_H:G\to G/H$ is the quotient map, and $\varphi_H:G/H\to B(X)$ is a new norm-continuous unitary representaion?

An equivalent formulation:

Let $\varphi:G\to B(X)$ be a norm-continuous unitary representation, and let $$ \operatorname{Ker}\varphi=\{g\in G:\ \varphi(g)=1\} $$ be its kernel. Is the quotient group $G/\operatorname{Ker}\varphi$ always a Lie group?

I thought that this follows from the results on the 5-th Hilbert problem, but when I tried to write the proof I understood that I can prove this only in the case when $G$ is commutative or compact.

I think there must be a trick that I don't know, or, on the contrary, there are some counterexamples. Can anybody enlighten me?

P.S. I understood that it is sufficient here to consider the case when the group $G$ is metrizable. But I'm stuck on this.

P.P.S. Perhaps, the following reformulation will simplify the question:

Can the group $U(X)$ of unitary operators on a Hilbert space $X$ (not necessarily separable) with the norm topology contain a closed subgroup $G$ which is locally compact, but not locally Euclidean (with respect to the topology induced from $U(X)$)?

Answer 1

I am sorry, I have realized that the answer is "yes", and this is simple. The proof is the following.

Suppose this is not true. Then we can find a locally compact group $G$ which is not locally Euclidean, but which can be (injectively) embedded $$ \pi:G\to U(X) $$ into the group $U(X)$ of unitary operators on a Hilbert space $X$ with the norm topology.

It is known (see V.M.Glushkov, 1957, 5.6, Theorem 9) that each locally compact group $G$ contains an open pro-Lie subgroup $G'$, i.e. a subgroup which can be represented as a projective limit of its quotient Lie groups: $$ G'=\projlim_{K\in\lambda(G')}G'/K $$ (here $\lambda(G')$ is the system of compact normal subgroups in $G'$ for which the quotient group $G'/K$ is a Lie group).

So we can think that $$ \pi:G\to U(X) $$ is a norm-continuous (and injective) unitary representation of a pro-Lie locally compact group $G$, in a Hilbert space $X$.

Now take a compact neighbourhood of unit $U$ in $G$, and choose a (compact) subgroup $K\subseteq U$ such that $K\in\lambda(G)$ (i.e. $G/K$ is a Lie group). The restriction $$ \pi\Big|_K:K\to U(X) $$ is again an embedding, but now this is an embedding of a compact group.

Now we decompose this norm-continuous representation into irreducible components, and we have (see A.I.Shtern, 2008, Corollary 2) that only finite number of isotypic components are non-trivial. In other words, this compact group $K$ can be embedded (injectively, continuously and homomorphically) into a direct sum of the form $$ B(X_1)\oplus B(X_2)\oplus ... \oplus B(X_n), \qquad n\in{\mathbb N}, $$ where $X_i$ are finite-dimensional Hilbert spaces.

This means that $K$ must be finite-dimentional, i.e. a Lie group.

So we have that both $G/K$ and $K$ are Lie groups. Hence, $G$ is also a Lie group. And this contradicts to our choice of $G$.