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.

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.