= Solution
Let $n=[G:H]<\infty$. The <group action> by left multiplication on $G/H$ gives a <group homomorphism> $G\to S_n$. Its kernel is the <normal core of a subgroup>,
$$
\boxed{N=\bigcap_{g\in G}gHg^{-1}\trianglelefteq G,\qquad N\leq H,\qquad[G:N]\leq n!.}
$$
The containment follows by looking at the stabilizer of the coset $H$, and finite index follows from the finite image in $S_n$.
The <Higman group> is
$$
J=\langle a,b,c,d\mid aba^{-1}=b^2,\ bcb^{-1}=c^2,\ cdc^{-1}=d^2,\ dad^{-1}=a^2\rangle.
$$
It is visibly a <finitely presented group>. To prove infinitude, first form
$$
K=\langle a,b,c\mid aba^{-1}=b^2,\ bcb^{-1}=c^2\rangle.
$$
It is the <amalgamated free product> of $\langle a,b\mid aba^{-1}=b^2\rangle$ and $\langle b,c\mid bcb^{-1}=c^2\rangle$, identifying their infinite cyclic subgroups generated by $b$. Each factor is an <HNN extension> of an <infinite cyclic group>, so its base and stable letter both have infinite order. No nonzero power of $a$ belongs to $\langle b\rangle$ in the first factor, by the map to $\mathbb Z$ sending $a$ to one and $b$ to zero. In the second factor, no nonzero power of $c$ belongs to $\langle b\rangle$: the stable-letter map forces a hypothetical equality $c^k=b^l$ to have $l=0$, and the base $c$ has infinite order. The <normal form theorem for an amalgamated free product> therefore shows that $\langle a,c\rangle\leq K$ is a rank-two <free group>.
Similarly,
$$
L=\langle c,d,a\mid cdc^{-1}=d^2,\ dad^{-1}=a^2\rangle
$$
contains $\langle c,a\rangle$ as a rank-two <free group>. Identifying these two free subgroups yields
$$
J=K*_{\langle a,c\rangle}L.
$$
The <normal form theorem for an amalgamated free product> embeds $K$ in $J$. In particular, $J$ contains a <free group> of rank two and is \b[infinite].
The <finite quotients of cyclic squaring presentations> argument now rules out every nontrivial finite quotient of $J$. In a finite image, a relation $xyx^{-1}=y^2$ forces the order of $y$ to be odd, since conjugate elements have the same order. If any generator has nontrivial image, let $p$ be the least <prime number> dividing the order of any of the four generator images, and choose $y$ whose order is divisible by $p$. Its predecessor $x$ conjugates it to its square. If $r$ is the order of $x$, iterating conjugation gives $2^r\equiv1\pmod p$. Hence the <multiplicative order> of $2$ modulo $p$ divides $r$. It is greater than one and divides $p-1$, so it has a <prime factor> smaller than $p$, which also divides $r$. This contradicts the minimal choice of $p$. Thus all four generator images are trivial. \b[$J$ has no nontrivial finite quotient], and the normal-core argument above implies that \b[$J$ has no proper finite-index subgroup].
For the final argument, <Conjugation> preserves the order of an element. Thus if one nonidentity element has finite order $n$, every nonidentity element has that same order, and $n\geq2$. Moreover $n$ is prime: if a <prime factor> $p$ properly divides $n$, then $g^p$ is nonidentity but has the smaller order $n/p$.
When $n\geq3$, the element $g^2$ is nonidentity, so choose $x$ with $xgx^{-1}=g^2$. The conjugator is not the identity, since $g^2\ne g$, and therefore $x^n=1$. Induction gives $x^kgx^{-k}=g^{2^k}$, and at $k=n$ this yields
$$
g=g^{2^n},\qquad n\mid2^n-1.
$$
But <Fermat's little theorem>, with the odd prime $n$, gives $2^n\equiv2\pmod n$, contradicting that divisibility.
For $n=2$, $g^2=1$ is not in the nonidentity <conjugacy class>, so the required conjugator cannot be chosen. Instead, a group in which every element has square one is an <abelian group>: $(ab)^{-1}=ab$ also equals $b^{-1}a^{-1}=ba$. In an <abelian group> every <conjugacy class> is a singleton, so one nonidentity class permits only one nonidentity element, giving a group of order two. This contradicts infinitude. \b[Consequently the infinite group in question is a $\boxed{\text{torsion-free group}}$.]
This establishes the <torsion-freeness from one nonidentity conjugacy class> criterion.
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