How to show that $\mathrm{ord}_m a = \mathrm{ord}_m \overline{a}$? Let $a \in Z$ and $m \in N$ such that $\gcd(a,m)=1$. How to show that $\mathrm{ord}_m a = \mathrm{ord}_m \overline{a}$, where $\overline{a}$ is the inverse of a modulo m?

Hint: Solution starts as follows: 
  $1 \equiv (a \overline{a})^{ord_m a} \equiv a^{ord_m a} \overline{a}^{ord_m a}\pmod m$...
  Problem: I don't understand why they don't just start with $1 \equiv a \overline{a}\pmod m$...

 A: HINT: $a^k\bar a^k=(a\bar a)^k$
Added: (That was written before you edited the question.) For any integer $k$ you have $$a^k\bar a^k=(a\bar a)^k=1^k=1\;.\tag{1}$$ Take $k=\operatorname{ord}_m(a)$: $(1)$ becomes $$\bar a^{\operatorname{ord}_ma}=a^{\operatorname{ord}_ma}\bar a^{\operatorname{ord}_ma}=1\;;\tag{2}$$ take $k=\operatorname{ord}_m\bar a$ instead, and it becomes $$a^{\operatorname{ord}_m\bar a}=a^{\operatorname{ord}_m\bar a}\bar a^{\operatorname{ord}_m\bar a}=1\;.\tag{3}$$
I expect that you know that if $a^n=1$, then $\operatorname{ord}_ma\mid n$, i.e., $n$ is a multiple of $\operatorname{ord}_ma$. (If not, that’s the first thing that you need to prove.) If you combine that fact with $(2)$ and $(3)$, it’s not hard to prove that $\operatorname{ord}_ma=\operatorname{ord}_m\bar a$.
A: Here is a tip: How are $\overline a^n$ and $\overline{a^n}$ related?
A: If you know some ring theory this should be pretty easy. Let us call the order of $a$ in the quotient $\Bbb{Z}/m\Bbb{Z}$ $d$ say. We can speak of the multiplicative inverse of $a$ in the quotient because $(a,m) = 1$. Also, you should  know that the order of $\overline{a}^{-1}$ must divide the order of $d$.
Suppose that there is $n < d$ such that $((\overline{a})^{-1})^n = 1$. Then.....
A: Let $b=\overline{a}$ and $n=ord_m(a)$. Then $1=ab$ implies $1=1^n=(ab)^n=a^n b^n = b^n$ and so $ord_m(b)\le n=ord_m(a)$. Since $a=\overline{b}$, we get $ord_m(a)\le ord_m(b)$ and so $ord_m(a)=ord_m(b)$.
A: Hint $\rm\ \ a\,\bar a = 1\ \Rightarrow\ a^n\,\bar a^n = 1\ \Rightarrow\ [\, \color{#C00}{a^n=1}\iff\color{green}{\bar a^n = 1}\,]$  
Hence $\rm\displaystyle\ ord\ a = \min\,\{n\in \mathbb N_1 :\color{#C00}{ a^n = 1}\} = min\,\{n\in \mathbb N_1 : \color{green}{\bar a^n = 1}\} = ord\ \bar a $
