That is, if $G= \langle a \, | \, a^m = e \rangle$, then $G$ is isomorphic to $\mathbb{Z_m}$.
2025-01-13 02:19:50.1736734790
How do I prove that a group with one generator and a single relation is isomorphic to $\mathbb{Z_m}$?
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You have a surjection from Z to G which sends 1 to the generator of the group, the kernel is a subgroup mZ of Z, thus G =Z/mZ