= Solution
The requested representation is the left <coset action>
$$
\rho:G\longrightarrow\operatorname{Sym}(G/H),\qquad
\rho(g)(xH)=gxH.
$$
It is well defined because equal <cosets> remain equal after multiplication by $g$, and $\rho(gg')=\rho(g)\rho(g')$. It is transitive. For $H=1$ it is the left regular action; for general $H$ it is the associated transitive <coset> representation.
Its kernel is the <normal core of a subgroup>
$$
N=\bigcap_{x\in G}xHx^{-1}.
$$
Indeed, $g$ fixes every $xH$ precisely when $x^{-1}gx\in H$ for every $x$. Thus $N\lhd G$ and $N\leq H$. If $[G:H]=i<\infty$, the image lies in the finite symmetric <group> $S_i$, so
$$
\boxed{[G:N]\leq i!,\qquad [H:N]=\frac{[G:N]}i\leq(i-1)!.}
$$
In particular $N$ has finite index in $H$.
For <Higman group as an amalgam of iterated HNN extensions>, use the cyclic squaring presentation
$$
\mathcal H=\langle a,b,c,d\mid
a^{-1}ba=b^2,\ b^{-1}cb=c^2,\ c^{-1}dc=d^2,\ d^{-1}ad=a^2\rangle.
$$
First form $L=\langle b,c\mid b^{-1}cb=c^2\rangle$, an <HNN extension> of the infinite <cyclic group> $\langle c\rangle$. Its base embeds, so $c$ has infinite order; $b$ has infinite order as well, since the <homomorphism> $L\to\mathbb Z$ sending $b\mapsto1,c\mapsto0$ is onto. Consequently $\langle b\rangle$ and $\langle b^2\rangle$ are isomorphic infinite cyclic <subgroups>. Adjoining the <stable letter> $a$ with $a^{-1}ba=b^2$ gives
$$
P=\langle a,b,c\mid a^{-1}ba=b^2,\ b^{-1}cb=c^2\rangle.
$$
We claim $\langle a,c\rangle$ is a rank-two <free group>. No nonzero power of $c$ belongs to $\langle b\rangle$: applying the map $b\mapsto1,c\mapsto0$ would make that power equal to $1$, contradicting the infinite order of $c$. Thus no nonzero power of $c$ belongs to $\langle b^2\rangle$ either. Any nonempty freely reduced word in $a,c$ that contains $a$ is therefore a reduced HNN sequence: between inverse $a$ letters, a nonzero $c$ power cannot create a pinch. <Britton's lemma> makes it nonidentity. A word containing only $c$ is nonidentity by the base embedding. This proves the claim.
Likewise
$$
Q=\langle c,d,a\mid c^{-1}dc=d^2,\ d^{-1}ad=a^2\rangle
$$
is obtained from $\langle d,a\mid d^{-1}ad=a^2\rangle$ by adjoining $c$, and its <subgroup> $\langle c,a\rangle$ is free of rank two by the same argument. Form the <amalgamated free product> $P*_{\langle a,c\rangle}Q$, identifying these two free <subgroups> generator by generator. Its presentation is exactly that of $\mathcal H$. The component embeddings allowed in the question show that it contains $P$, hence is infinite. It has four <group generators> and four relators, so it is finitely presented.
To show that this <Higman group> has no nontrivial finite quotient, consider any finite image and the orders of the four generator images. A relation $x^{-1}yx=y^2$ makes $y$ and $y^2$ have the same order, so every generator order is odd. Suppose one is nontrivial, and let $p$ be the smallest prime dividing any of the four orders. Choose a generator $y$ whose order is divisible by $p$, with predecessor $x$ satisfying the displayed relation. Iterating conjugation $|x|$ times gives
$$
y^{2^{|x|}}=y,\qquad 2^{|x|}\equiv1\pmod p.
$$
The <multiplicative order> $d$ of $2$ modulo $p$ thus divides $|x|$. Also $d\mid p-1$ and $d>1$, since $2\not\equiv1\pmod p$. Any prime divisor of $d$ is smaller than $p$ and divides the order of $x$, contradicting minimality of $p$. Hence every generator image is trivial, so every finite image is trivial. A proper finite-index <subgroup> would give a nontrivial transitive finite <coset> image, which is impossible. Therefore \b[$\mathcal H$ is infinite and finitely presented, with no proper finite-index <subgroups>].
A <maximal normal subgroup> is a proper <normal subgroup> $M\lhd G$ with no <normal subgroup> strictly between $M$ and $G$. To prove that <finitely generated groups have maximal proper normal subgroups>, fix a proper <normal subgroup> $K$ and order the proper <normal subgroups> containing $K$ by inclusion. For a chain, its union is normal and contains $K$. It is still proper: if the union were all of $G$, each element of a finite generating set would lie in some chain member, and the chain order would put all these finitely many <group generators> in one member. That member would equal $G$, a contradiction. The <Zorn lemma> therefore supplies a maximal proper <normal subgroup> $M$ containing $K$.
Apply this to $\mathcal H$ and $K=1$. By the normal-subgroup correspondence, $\mathcal H/M$ is nontrivial and simple. It is finitely generated as a quotient of $\mathcal H$. It cannot be finite, since $\mathcal H$ has no nontrivial finite quotients. Thus
$$
\boxed{\text{there exists an infinite finitely generated simple group}.}
$$
The final assertion for arbitrary proper <subgroups> is false. Take $G=S_3$ and $H=\langle(12)\rangle$. If a <normal subgroup> contained $H$, it would contain all conjugate transpositions, which generate $S_3$, so it would equal $G$. Hence this proper <subgroup> lies in no proper <normal subgroup>, and in particular in no maximal <normal subgroup>. The maximal-normal existence result requires the starting <subgroup> itself to be normal.
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