the connection between $\gamma_m(a)$ and $\gamma_m(b)$ when $a\cdot b\equiv 1\pmod m$ 
show the connection between the order of $a$ $\gamma_m(a)$ and the order of $b$ $\gamma_m(b)$ when
$$a\cdot b\equiv 1\pmod m$$

I took $a=5$ and $b=4$
$$5\cdot 4\equiv 1\pmod{19}$$
$$\gamma_m(a)=9\text{ and } \gamma_m(b)=9$$
So I think that alaways $\gamma_m(a)=\gamma(b)$ when $a\cdot b\equiv 1\pmod m$
Is there a more formal way to show this?

$a$ and $b$ are each other's multiplicative inverse, i.e. $b=a^{-1} \mod m$. So the equation is $a\cdot a^{-1} = 1 \mod m$. Clearly, if you raise $1$ to a power, it remains $1$, so the same should be true of $a\cdot a^{-1}$. Suppose the order of $a$ is $\gamma(a)$, so $a^{\gamma(a)}=1\mod m$. Then
$$1 = a\cdot a^{-1} = (a\cdot a^{-1})^{\gamma(a)} = (a)^{\gamma(a)}\cdot (a^{-1})^{\gamma(a)} = 1 \cdot (a^{-1})^{\gamma(a)} = (a^{-1})^{\gamma(a)} = 1$$
So the orders are equal. $\square$
