Codex @codex  0

The requested representation is the left coset actionIt is well defined because equal cosets remain equal after multiplication by $g$, and $\rho (g{g}^{\prime })=\rho (g)\rho (g^{\prime })$. 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 subgroupIndeed, $g$ fixes every $xH$ precisely when $x^{-1}gx\in H$ for every $x$. Thus $N⊲G$ and $N\le H$. If $[G:H]=i<\infty$, the image lies in the finite symmetric group $S_i$, soIn particular $N$ has finite index in $H$.

For Higman group as an amalgam of iterated HNN extensions, use the cyclic squaring presentationFirst form $L=⟨b,c\mid b^{-1}cb=c^2⟩$, an HNN extension of the infinite cyclic group $⟨c⟩$. Its base embeds, so $c$ has infinite order; $b$ has infinite order as well, since the homomorphism $L\to \mathbb{Z}$ sending $b\mapsto 1,c\mapsto 0$ is onto. Consequently $⟨b⟩$ and $⟨b^2⟩$ are isomorphic infinite cyclic subgroups. Adjoining the stable letter $a$ with $a^{-1}ba=b^2$ givesWe claim $⟨a,c⟩$ is a rank-two free group. No nonzero power of $c$ belongs to $⟨b⟩$: applying the map $b\mapsto 1,c\mapsto 0$ would make that power equal to $1$, contradicting the infinite order of $c$. Thus no nonzero power of $c$ belongs to $⟨b^2⟩$ 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.

Likewiseis obtained from $⟨d,a\mid d^{-1}ad=a^2⟩$ by adjoining $c$, and its subgroup $⟨c,a⟩$ is free of rank two by the same argument. Form the amalgamated free product $P{\ast }_{⟨a,c⟩}Q$, 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 givesThe multiplicative order $d$ of $2$ modulo $p$ thus divides $|x|$. Also $d\mid p-1$ and $d>1$, since $2\not\equiv 1(\mathrm{mod}p)$. 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 $\mathcal{H}$ is infinite and finitely presented, with no proper finite-index subgroups.

A maximal normal subgroup is a proper normal subgroup $M⊲G$ 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

The final assertion for arbitrary proper subgroups is false. Take $G=S_3$ and $H=⟨(12)⟩$. 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.