Yes. Non-open maximal subgroups are not closed and their closure is equal to the whole of the group. I mention above some words about the motivation

What I ask is equivalent to find a non- normal maximal subgroup

I think you are right. But I thought to find a non-normal maximal subgroup.

@YCor: I did not (couldn't maybe better to say) compute the kernel of $G\rightarrow \tilde{G}$. Instead as I mentioned in the question, I was unable to find the abelianization of $G^{(4)}$, where its index in $G$ is $19730006016$.

Many Thanks. It remains to decide if it is infinite or not.

Could you please give a reference for your question?

@BhaskarVashishth: You must quote that $G$ is "not" torsion-free.

Again to record, for any two numbers $m,n$ (not both zero, of course) we have $(1+x+y)( n(x+x^h+\cdots+x^{h^6})+m(x^{-1}+y+y^h+\cdots+y^{h^5})-(n+m)(1+h+\cdots+h^6) )=0$

Using the above answer let me give an element $\beta$ of $\mathbb{Q}[G]$ such that $(1+x+y)\beta=0$ to record one. Let $G=\langle a,h \;|\; a^3=h^7=1, h^a=h^4 \rangle$. Now take $x:=a$ and $y:=(a^h)^{-1}$. Then $(1+x+y)(x+x^h+\cdots+x^{h^6}-x^{-1}-y-y^{h}-\cdots-y^{h^5})=0$. The support of $\beta$ has size $14$. There is another $\beta$ with support of size $21$.

Very nice. Many thanks. Have you another example in which the orders of $x$ and $y$ are not $3$? One key in your example is that $(1+x+x^2)(1-x)=0$ if $x^3=1$. Anyway, thanks again.