That seems like it would make sense, but I can't think of any reasoning behind it. Could anyone tell me if it's the case, and if it is how can i justify it?
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Follows immediately by the pigeonhole principle - see the Lemma in the linked dupe. – Bill Dubuque Nov 27 '22 at 13:56
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If you already know group theory then note that the order of $a$ is the size of the subgroup generated by $a$ which is clearly $\le $ the size of the whole group (which here is $\le n-\color{#c00}1$ since it omits nonunit $\color{#c00}0$ in $\Bbb Z_n)\ \ $ – Bill Dubuque Nov 27 '22 at 14:01
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It is. The order of the multiplicative group modulo $n$ is $\varphi (n)\le n-1$. The order of an element can't exceed the order of the group.
The upper bound is optimal as $\varphi (p)=p-1$ for $p$ prime, and in this case the group is cyclic.
calc ll
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Please strive not to post more (dupe) answers to dupes of FAQs (and PSQs), cf. recent site policy announcement here. – Bill Dubuque Nov 27 '22 at 13:56