If ${\phi}_m : C_n \rightarrow C_n$ is defined by ${\phi}_m(x^t) = x^{mt}$, what is a necessary and sufficient condition on m for ${\phi}_m$ to be an automorphism?
Thanks in advance.
What is the condition required for homomorphism ${\phi}_m : C_n \rightarrow C_n$ to be an automorphism?
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abstract-algebra
group-theory
group-homomorphism
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0$(m, n) = 1$ I think? – 2017-01-10
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0@q.Then is right. What would happen otherwise? – 2017-01-10
1 Answers
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Let $a$ denote the generator of $C_n$ so $|a| = n$. Then we have a canonical way to view $\phi_m$ as: $$\phi_m(a^t) = x^{mt}$$ Where $t \in \mathbb{N}$ such that $1 \leq t \leq n$. First, note that $(m, n) = 1$, then really, the map takes $t$ and sends it to $tm \pmod n$. So we are really considering $m$ as an element in $\mathbb{Z}_n$. If we want an automorphism, we require that $m, 2m, 3m, ..., mn \pmod n$ produce all $n$ values, that is precisely that $|m| = n$ in $C_n \approx \mathbb{Z}_n$. What condition is required for $|m| = n$ in $\mathbb{Z}_n \approx C_n$?
If $(m, n) \neq 1$, then we have a common divisor $d$ between them. The proof is extremely similar in showing that if we have a group $\mathbb{Z}_n$, then $x \in \mathbb{Z}_n$ has order $n$ if and only if $(x, n) = 1$.
Hint: Look at the gcd of elements in $\mathbb{Z}_n$ against $n$.