abstract algebra - Proof of $(mathbb{Z}/p^k mathbb{Z})^timescongmathbb{Z}/phi(p^k) mathbb{Z}$

While learning some material about primitive roots, I read some algebraic approach of the proof of the existence of primitive roots.

I read this wikipedia article, however, I got a little bit confused with this following statement (why it is cyclic?) :

For each odd prime $p^k$, the corresponding $(\mathbb{Z}/p^k \mathbb{Z})^\times$ is a cyclic group of order $\phi(p^k)= p^k - p^{k-1} $, which may further factor into cyclic groups of prime-power orders.

I tried to prove that the order of $\bar 2$ in this group is $\phi(p^k)$, but didn’t make any progress.

Please help.

Answer

$\overline{2}$ is not even a generator of $(\mathbb{Z}/p\mathbb{Z})^{\times}$ for each odd prime $p$. In the case $p=7$, we have $$2^3=8\equiv 1\mod 7.$$
Therefore, the order of $\overline{2}$ in $(\mathbb{Z}/7\mathbb{Z})^{\times}$ is $3$, whereas $|(\mathbb{Z}/7\mathbb{Z})^{\times}|=6$.