$ord(a,b)$ is the order of the element (a,b) in the additive group $(\mathbb{Z}/(n))^2$.
Why is $ord(a,b)\cdot gcd(a,b,n)=n$?
Thank you in advance.
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The order of $(a,b)$ in $(\mathbb Z_n)^2$ is the smallest $k$ such that $$ k(a,b)=(0,0)\in(\mathbb Z_n)^2\iff ka=kb=0\in\mathbb Z_n $$ Now let $k$ be any integer satisfying the above relation. Then using Bezout's identity $$ \gcd(a,b)=\lambda a+\gamma b $$ and considering it modulo $n$ we get $$ k\cdot\gcd(a,b)=\lambda ka+\gamma kb=\lambda 0+\gamma 0\in\mathbb Z_n $$ In particular this works for $k=\operatorname{ord}(a,b)$ which is just a particular such value of $k$.
String
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2Thank you @String but why ist $ord(a,b)\cdot gcd(a,b,n)=\lambda ka+\gamma kb+\alpha kn=n?$ and not a multiple of $n$? – Mandarine Mar 05 '17 at 13:15
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The order of $(a,b)$ in $\mathbb{Z}_n \times \mathbb{Z}_n$ is given by: $LCM\left(\frac{n}{(a,n)}, \frac{n}{(b,n)}\right)$. Now we have:
$$\left[\frac{n}{(a,n)}, \frac{n}{(b,n)}\right] = \frac{n^2}{(b,n)(a,n)\left(\frac{n}{(a,n)}, \frac{n}{(b,n)}\right)} = \frac{n^2}{(b,n)(a,n)\frac{n}{[(a,n),(b,n)]}} = \frac{n(a,n)(b,n)}{(b,n)(a,n)((a,n),(b,n))} = \frac{n}{(a,b,n)}$$
Where we used the notation $(a,b) = GCD(a,b)$ and $[a,b] = LCM(a,b)$
Stefan4024
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$n\mid ka,kb \iff n\mid ka,kb,kn\!\! \overset{\ \ \rm\color{#c00}{U}}\iff n\mid (ka,kb,kn)\overset{\rm\color{#0a0}{D}}=(a,b,n)k\iff n/(a,b,n)\mid k$
${\rm ord}(a,b)\, $ is the least positive such $k,\,$ which, by above, is clearly $\,k = n/(a,b,n).$
Above we used $\:\rm\color{#c00}{U} = $ gcd Universal property: $\,\ n\mid i,j\iff n\mid(i,j)$
Also, $ $ we used $\, $ $\rm\color{#0a0}{D}$ = gcd Distributive law: $\,\ (ik,jk,\ell k) = (i,j,\ell)k$
Bill Dubuque
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@Mandarine Wrong, because in the final arrow we require $,(a,b,n)\mid n,$ when we cancel $,(a,b,n),$ from the divisibility relation. But that won't work for $,(a,b),$ since generally $,(a,b)\nmid n\ \ $ – Bill Dubuque Mar 05 '17 at 14:58
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