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?

One has $a^{\gamma_m(a)}\equiv 1$ and $b^{\gamma_m(b)}\equiv 1\pmod{m}$. Multiplying the two and assuming without loss of generality that $\gamma_m{a}\gt \gamma_m(b)$ we can write
$$a^{\gamma_m(a)-\gamma_m(b)}\left(a\cdot b\right)^{\gamma_m(b)}\equiv 1\pmod{m}$$
This means $a^\alpha\equiv 1\pmod{m}$ with $\gamma_m(a)-\gamma_m(b)=\alpha\lt \gamma_m(a)$ which is a contradiction. So the orders are equal.
