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If $(a,m)=(b,m)=1$ and if $(\exp_m(a),\exp_m(b))=1$, prove that $$\exp_m(ab)=\exp_m(a)\exp_m(b).$$

The notation $\exp_m(a)$ is denote the smallest positive integer $n$ such that $a^n\equiv 1\pmod m$.

My proof: Let $f=\exp_m(a)$, $h=\exp_m(b)$, then let $n=\exp_m(ab)$, we have $$(ab)^n\equiv 1\pmod m$$ I want to argue that $f|n, h|n$ so $n=fk$, but I realize it is also possible that there is $f'|f$ and $h'|h$, although $a^{f'}\not\equiv 1\pmod m$, and $b^{h'}\not\equiv 1\pmod m$, but we can also have $a^{f'}b^{h'}\equiv 1\pmod p$, I cannot disprove this possibility, any suggestion?

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Suppose, by way of contradiction, that it is possible that $$ a^{f'}b^{h'}\equiv 1\pmod p. $$ Since $f'\vert f$, the ratio $f/f'$ is an integer. Raise the congruence to that power. Since $a^f\equiv 1\pmod p$, this reduces to $$ b^{h'{f}/{f'}} \equiv 1 \pmod p. $$ Since $h'$ is a proper divisor of $h$ and $f$ is coprime to $h$, the exponent $h'f/f'$ is not a multiple of $h$. However, since $h$ is the smallest exponent for which $b^h\equiv1\pmod p$, every such exponent must be a multiple of $h$. So we have a contradiction.

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  • $\begingroup$ Very useful! The part raising to the $f/f'$ power is the point I haven't done. Thank you! $\endgroup$ – kelvin hong 方 Mar 13 at 5:44

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