Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $1\le d$, $1 \le m$ where $d \mid m$. Suppose that $\gcd(a,m)=1$ (and so $\gcd(a,d)=1$). Prove that $\mathrm{ord}_d(a)$ divides $\mathrm{ord}_m(a)$.

share|improve this question

2 Answers 2

Let $x=\mathrm{ord}_m(a)$. Then we have $$a^x\equiv 1 \pmod m\implies a^x=mk+1$$ for some $k\in\mathbb{Z}$. Let $m=m'd$. Then $$a^x = mk +1 = d(m'k)+1$$ What can you conclude from this?

share|improve this answer

$a^{ord_d(a)}\equiv 1\pmod d$

Let $a^h \equiv 1\pmod d$ and $ord_d(a)<h=c \cdot ord_d(a) +r$ where $0\le r<ord_d(a)$

So, $a^{c \cdot ord_d(a) +r}\equiv 1\implies a^r\cdot (a^{ord_d(a)})^c\equiv 1$ $\implies a^r\equiv 1$,

So there is one value $r$ of $x$ which $<ord_d(a)$ to satisfy $a^x\equiv 1$, but we know, $ord_d(a)$ is the smallest positive integer value of $x$ , so $r=0\implies ord_d(a)\mid h$

$a^{ord_m(a)}\equiv 1\pmod m$

$\implies a^{ord_m(a)}\equiv 1\pmod d$ as $d\mid m$

$\implies ord_d(a)\mid ord_m(a)$ as here $h=ord_m(a)$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.