Higman group ($Q$)

The Higman group has the finite group presentation $Q=⟨a_1,a_2,a_3,a_4\mid a_i{a}_{i+1}{a}_i^{-1}=a_{i+1}^2(i\mathrm{mod}4)⟩$. Its subgroup $⟨a_1,a_3⟩$ is a rank-two free group, and the normal closure of each $a_i$ is all of $Q$. It has no nontrivial finite quotients of a group: a least prime divisor among the four generator orders in a finite image contradicts the multiplicative-order constraints imposed by conjugation to squares. These properties give finite-presentation constructions embedding arbitrary finitely presented groups into groups with no nontrivial finite quotients.

The two three-generator components on $\{a,b,c\}$ and $\{c,d,a\}$ are iterated HNN extensions of infinite cyclic groups. The subgroup on $\{a,c\}$ is free of rank two in each component by Britton's lemma. Amalgamating over these free subgroups gives the displayed infinite finitely presented group. Its cyclic squaring relations forbid every nontrivial finite quotient by the least-prime multiplicative-order argument.

A group presentation with a cyclic list of generators $a_i$ and relations $a_i{a}_{i+1}{a}_i^{-1}=a_{i+1}^2$ has no nontrivial finite quotient of a group. In a finite image, all generator orders are odd. Choose the least prime $p$ dividing any nontrivial generator order and let $x$ be the predecessor of a generator $y$ whose order is divisible by $p$. Conjugation by $x$ acts as squaring on $⟨y⟩$, so the multiplicative order of $2$ modulo $p$ divides the order of $x$. That multiplicative order is greater than one and divides $p-1$, giving a prime divisor smaller than $p$ in the order of $x$, a contradiction. This argument applies to any cycle length; it asserts absence of finite quotients, not infinitude of the presented group.