Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 133 / 1 / a

Use the Schreier coset graph of the subgroup, with one directed edge labelled $s$ from $Hg$ to $Hgs$ for each $s\in \{a,b\}$. The homomorphismidentifies the cosets of $\ker \varphi$ with the integers. The covering graph therefore has vertices $v_n$, $n\in \mathbb{Z}$, an $a$-edgeand a $b$-loop at every $v_n$. Thus it is a doubly infinite $a$-line with one $b$-circle attached at each integer vertex.

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Forthe map $\varphi :F_2\to \mathbb{Z}$ is surjective because $\gcd (2,3)=1$. Again identify the cosets of $\ker \varphi$ with $\mathbb{Z}$. The covering has vertices $v_n$ and directed edgesThis labelled graph is connected: integer combinations of $2$ and $3$ reach every vertex. Each vertex has one incoming and one outgoing edge of each label, as required for a covering graph of the two-petalled rose.

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The based Schreier coset graph has vertices $⟨a⟩g$ and an $s$-edge from $⟨a⟩g$ to $⟨a⟩gs$. At the base vertex $⟨a⟩$ there is an $a$-loop. There are no other cycles: the fundamental group of the covering is the subgroup $⟨a⟩\cong \mathbb{Z}$, and its displayed loop already generates it.

Equivalently, start with one $a$-circle at the base vertex and attach labelled trees so that every vertex has exactly one incoming and one outgoing $a$-edge and exactly one incoming and one outgoing $b$-edge. After deleting the base $a$-loop, the underlying graph is a tree. This describes the complete infinite covering and distinguishes it from its finite core, which is just the $a$-loop.

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