Topological group: if $g_n, h_n, k_n \in G$ s.t. $g_n h_n \to e$ and $k_n \to e$ does $g_n k_n h_n \to e$?

Question

This is certainly true if $g_n$ (equivalently $h_n$) converge in $G$ and it feels like it should be true but after trying to prove it, it's unclear.

Answer 1

I think this should give a counterexample: set $G := GL_2(\mathbb{R})$, and $$g_n := \begin{bmatrix} n & 0 \\ 0 & 1 \end{bmatrix}, k_n := \begin{bmatrix} 1 & 1/n \\ 0 & 1 \end{bmatrix}, h_n := \begin{bmatrix} 1/n & 0 \\ 0 & 1 \end{bmatrix}.$$ Then for each $n$, $g_n k_n h_n = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$.