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Theorem: If $ord_{m}a=t$, then $ord_{m}a^n=t/(n,t)$

Proof: Let $(n,t)=d$. Then since $a^t \equiv 1(mod\text{ } m)$, we have

$$(a^t)^{n/d}=(a^n)^{t/d} \equiv 1(mod\text{ } m)$$,

so that if $ord_{m}a^n = t'$, then $$t'|t/d$$

But from the congruence $$(a^n)^t{'} \equiv 1(mod \text{ } m)$$

We have $t|nt'$ by theorem 4-3, or

$$t/d|{n/d}t'$$

Since

$$(t/d,n/d)=1$$

we obtain $$t/d|t'$$

Combining $t'|t/d$ and $t/d|t'$ we have $t'=t/d$

why do we need n/d in the first statement and how does that show $t'|t/d$

TheMathNoob
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  • See http://archive.org/details/NumberTheory_862 @Page#95. See also : http://math.stackexchange.com/questions/1815832/find-all-the-primitive-roots-of-13 – lab bhattacharjee Jul 11 '16 at 05:02

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The argument is essentiallyas follows (reorganized and presented bidirectionally)

$$ a^{nk}\equiv 1\iff t\mid nk\iff t\mid nk,tk\iff t\mid(nk,tk)=(n,t)k\iff t/(n,t)\mid k$$

The first $\iff$ follows because $\,t = {\rm\ ord}\, a,\,$ and the third follows by the definition/universal property of the gcd and the gcd distributive law. You could also use lcm instead of gcd, i.e.

$\quad\ \ a^{nk}\equiv 1\iff t\mid nk\iff n,t\mid nk\iff [n,t]\mid nk\iff [n,t]/n\mid k$

Both are equivalent since $\ [n,t]/n = t/(n,t),\ $ i.e. $\ [n,t](n,t) = nt$.

Remark $\ $ Note that the first equivalence chain implies that $\,a^n\,$ has order $\,t/(n,t)\,$ since, generally, if $\ b^k = 1\iff j\mid k\ $ then this implies that $\,b\,$ has order $\,j.\,$ Indeed, setting $\,k=j\,$ implies $\,b^j=1\,$ and $\,j\,$ is the least postive $\,k\,$ with $\,b^k=1\,$ since $\,j\,$ divides all other such $\,k.\ $

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Bill Dubuque
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  • Hi Bill Dubuque, I don't know why, but most of the times, it's so hard for me to follow your arguments. I still lack a lot of math background. I am just wondering why the $gcd(t/d,n/d)=1$ – TheMathNoob Jul 11 '16 at 04:02
  • If $,c>1,$ and $,c\mid t/d,n/d,$ then $,cd\mid t,n,,$ contra $,d = (t,n)=,$ greatest common divisor of $,t,n., $ See this answer.You are quite welcome to ask questions about things you find hard to follow. – Bill Dubuque Jul 11 '16 at 04:18
  • Thanks, I think you are using new notation because in mine $ord_m{a^n}=t'$. In yours, it is t. what is $t|nk,tk$?, are u saying that $t$ divides $nk$ and$ tk$? – TheMathNoob Jul 11 '16 at 04:24
  • No, same notation. The proof above shows that $,a^{n,k}=1\iff t_1\mid k,,$ for $,t_1 = t/(n,k).,$ This implies $,t_1 = {\rm ord}, a^{n}, (=: t')$ since for $,k=t'$ it implies $,t_1\mid t'.,$ Conversely for $,k=t_1,$ it implies $,(a^n)^{t_1} = 1,$ so $,t'\mid t_1.,$ Thus $,t_1\mid t'\mid t_1,$ so $,t' = t_1.\ $ That $\iff$ is a very handy characterization of "order". – Bill Dubuque Jul 11 '16 at 04:52
  • @TheMathNoob Some of your difficulties may be from proceeding too fast. Be sure you have properly understood things before moving forward. – Bill Dubuque Jul 11 '16 at 04:57