Consider all bijective mappings from $G \to G$ ($G$ is a finite cyclic group) in which generators are mapped only to generators. Are all these mappings isomorphisms?
I know that an isomorphism between $2$ cyclic groups takes a generator to generator. I want to know whether the reverse is true as well if given that the mapping is a bijection.
Best Answer
No. Consider $G = \mathbb{Z} / 5 \mathbb{Z}$. Every element except $0$ is a generator. There are only $4$ automorphisms of $G$ since a group homomorphism $G \to G$ is entirely determined by where it sends $1$, since $1$ generates $G$, and you can’t send $1$ to $0$. In other words, the only automorphisms of $G$ are the ones that multiply by $1, 2, 3,$ or $4$.
But there are $4! = 24$ different permutations of the generators $\{ 1, 2, 3, 4 \}$. So there are $24-4=20$ counterexamples.
EDIT: In general, if $p$ is prime, then $\mathbb{Z} / p \mathbb{Z}$ has $(p - 1)$ generators, has $(p - 1)$ automorphisms, and has $(p - 1)!$ permutations of its generators.