I don't think so, because my intuition tells me that if d is an isomorphism from H to G and that a generates H, then d(a) must generate G. However, the Z4 (cyclic) is isomorphic to U(8) (non-cyclic). What is going on here ?
2026-04-07 09:41:41.1775554901
Can a non-cyclic group be isomorphic to a cyclic group and vice-versa?
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No, cyclicity is preserved by isomorphism.
There's only one cyclic group of given finite order; and one infinite one. Cyclic groups are singly generated. That means that there is at least one element whose order is the order of the group.
And, $U(8)\cong V_4$ is not cyclic, but is rather the Klein four group. You can check that $\pmod8$, none of $1,3,5,7$ have order $4$.
Indeed, $1\equiv 1,\,3^2\equiv 1,\,5^2\equiv 1$ and finally $7^2\equiv 1$.