Does the order of $2$ in $Z_p$ divide the order of $2$ in $Z_n$?

Question

Givens. Consider $(Z_n, \times)$ the group of integers coprime with $n = pq$, where $p, q$ are prime numbers. Similarly, $(Z_p, \times)$ is the group of integers $\{0, 1, 2, ..., p - 1\}$ coprime with $p$.

Question. Is it true that $o(2, p)$ divides $o(2, n)$, where $o(i, y)$ describes the order of $i$ in $(Z_y, \times)$? Why?

Remarks. I know the size of the group $Z_n$ is $\varphi(n) = (p -1)(q - 1)$, the Euler totient function. I also know that $o(2, p) = p$ because $p$ is prime and every group of prime order is cyclic and any member of it is a generator --- I can prove that. But I don't really see if these facts I have are relevant to the question.

Answer 1

If $2^k\equiv 1\mod{n}$, then also $2^k\equiv 1\mod{p}$, so that $k$ is a multiple of $o(2,p)$.

This applies in general: if $m\mid n$, then $o(i,m)\mid o(i,n)$ (as long as $i$ is prime to both $m$ and $n$).

Answer 2

Yes. In fact we can establish something more general. Let $r$ be any positive integer and let $p_1,\ldots, p_r$ be any $r$ primes. Next let $a$ an integer. Suppose $a^l \equiv 1$ $\mod p_1p_2\ldots p_r$ (and note that $a$ does not have to be 2). Then $a^l = 1 + m(p_1 \ldots p_r)$ $=1 +(mp_2p_3 \ldots)p_1$ which implies that $a^l \equiv 1$ mod $p_1$.

Answer 3

Yes. More generally:

If $\phi: G \to G'$ is a group homomorphism and $g \in G$ has finite order, then $o(\phi(g))$ divides $o(g)$.

See a proof here.

This general fact applies to your case because $x \bmod n \mapsto x \bmod d$ defines a group homomorphism $\mathbb Z_n^\times \to \mathbb Z_d^\times$ if $d$ divides $n$.