= Finite quotients of cyclic squaring presentations
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 $\langle y\rangle$, 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.
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