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I was reading a proof (from here, Theorem 22) which used the following statement

"... for odd prime p ... if g is a primitive root mod p, then [Handout A2.5, Corollary 2] the order of g mod $p^a$ is divisible by p−1, hence of the form $p^k\cdot(p − 1)$ for some $k \leq a − 1$. ..."

It can be observed that $g^i \equiv 1 (mod \space p^{a}) \implies g^i \equiv 1 (mod \space p)$

So multiple passes over the elements of the group mod p are required in order to arrive at $1\space mod \space p^{a}$.

It is also known that the order of the multiplicative group of integers modulo $p^a$ is $p^{a-1}(p-1)$ and the order of a subgroup divides the order of the group.

Considering the observations above, indeed, the order of g mod $p^a$ is of the form $p^k\cdot(p − 1)$ for some $k \leq a − 1$.

But the corollary that it is referred states:

"Corollary 2. If G and G′ are finite groups such that there exists a surjective group homomorphism f : G → G′, then #G′ | #G."

I cannot find how this corollary can be applied to this problem.

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If $G = (\mathbb{Z}/p^a\mathbb{Z})^{\times}$ and $G' = (\mathbb{Z}/p\mathbb{Z})^{\times}$ then reduction mod $p$ gives a surjective group homomorphism $f : G \rightarrow G'$ and so $|G'|$ divides $|G|$ (but we knew that already, right?) and the rest of the argument would proceed exactly as you proved it.

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  • $\begingroup$ But that corollary only tells us that $ord(G') = K\cdot(p-1)$. I don't think this information alone is sufficient to prove that $ord(g)$ is a multiple of $p-1$. $\endgroup$
    – user42768
    May 2 '17 at 20:07

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