Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 120 / 1 / a

Use the synchronous group multiplier automata belonging to the automatic structure for a group. Write $u\otimes v$ for the padded convolution of words: the shorter tape is completed with a symbol $\#$ outside the alphabet, and reading stops when both words over an alphabet have ended. Our convention isLet $N\ge 1$ be at least the number of states of each of these finite-state automata, including $M_{\epsilon }$.

Choose a shortest representative $v\in L$ of $g$. In the first orientation, $M_x$ accepts $w\otimes v$; in the other orientation it accepts $v\otimes w$. If $|v|>|w|+N$, the accepting run has more than $N$ transitions after the tape containing $w$ has ended. Two states in that suffix coincide. Delete the intervening nonempty loop. Every deleted column contains $\#$ on the fixed tape and a genuine letter on the tape containing $v$, so the resulting input remains a valid padded convolution of words. The group multiplier automaton still accepts, giving a shorter representative of the same $g$. This contradicts minimality. Thus a representative can always be chosen withThis argument does not require a symmetric alphabet: the second orientation uses the same $M_x$ with its tapes interchanged in the input, rather than a presumed $M_{x^{-1}}$.

Apply the same pumping of a group multiplier automaton to the accepted pair containing the given representative $u$. Since $|u|>|w|+N$, there is a nonempty loop entirely after the tape containing $w$ has ended. Write $u=pqr$, with $q\ne \epsilon$ the letters read during this loop. Repeating the loop any number of times leaves the other tape equal to $w$, and the group multiplier automaton accepts the resulting padded convolution of words. ConsequentlyThese words over an alphabet have distinct lengths because $|q|>0$. Hence there are infinitely many representatives of $g$ in $L$. No uniqueness of representatives was assumed in either part.