From this question , its proved that for all co-primes $a$ of $n(=pq)$ ,
$a^\lambda \equiv 1 \mod n$
where $\lambda= lcm (p-1,q-1)$
But how to prove that it is the smallest one possible .
My argument : If there exists smallest $x(<lcm(p-1,q-1))$ such that $a^x \equiv 1 \mod n$ for all co-prime $a$ of $n$.Then either $(p-1)\not | x$ or $(q-1)\not | x$ . So either $a^x \not \equiv 1 \mod p$ or $a^x \not \equiv 1 \mod q$ , and tus we cant combine as $a^x \equiv 1 \mod pq$
Is my argument valid/correct ?
We assume $p$ and $q$ are distinct primes.
There is an object $a$ such that $p-1$ is the smallest positive $k$ such that $a^k\equiv 1\pmod{p}$. Such an $a$ is called a primitive root of $p$. Similarly, let $b$ be a primitive root of $q$.
By the Chinese Remainder Theorem, there an $x$ congruent to $a$ modulo $p$, and congruent to $b$ modulo $q$.
If $k$ is a positive integer such that $x^k\equiv 1\pmod{pq}$, then the congruence holds modulo $p$. But $x$ is a primitive root of $p$, since it is congruent to $a$. So $p-1$ divides $k$. Similarly, $q-1$ divides $k$, So $k$ is a multiple of the lcm of $p-1$ and $q-1$. In particular, it is greater than or equal to that lcm.
As you observed, the lcm of $p-1$ and $q-1$ works, so it is the smallest positive integer that does.
Remark: The argument given in the post is not clear, and it is at least quite incomplete. A CRT argument is needed to produce an $x$ which is simultaneously of largest possible order modulo both $p$ and $q$.