Cyclic $\mathbb{Z}_{2p^k}^*$ group

I'm new in group theory and have some problems.

I have got difficulties with understanding my notes from lecture. Maybe you can help me? I have this proof that $\mathbb{Z}_{2p^k}^*$ ($p>2$ prime) is a cyclic group. It uses the fact that $\mathbb{Z}_{2p^k}^* \simeq \mathbb{Z}_ 2^* \oplus \mathbb{Z}^*_{p_k}$ and both are cyclic groups, so it sufficies that we find odd generator of $\mathbb{Z}^*_{p^k}$, but why exactly? What does it give us that we have pair of generators of above groups, how we construct generator of $\mathbb{Z}_{2p^k}^*$ from them?

The second problem I don't understand is that we can always find an odd generator of $\mathbb{Z}^*_p$. Clearly it is a cyclic group so there exists some generator $g$. If it is odd then we are done, but if not, then we take $p^k+g$ which is odd. It is also a generator of $\mathbb{Z}^*_p$, which is obvious but is it also a generator of $\mathbb{Z}^*_{2p^k}$? I was trying to prove that, but nothing came to my mind.

I really want to understand this approach. Can anybody help me?

• Since $\mathbb{Z}_2^*\simeq\mathbb{Z}_1$, we can simplify $\mathbb{Z}_{2p^k}^* \simeq \mathbb{Z}_ 2^* \oplus \mathbb{Z}^*_{p_k}$ to $\mathbb{Z}_{2p^k}^* \simeq \mathbb{Z}^*_{p_k}$, so an odd generator in $\mathbb{Z}^*_{p_k}$ is a generator in $\mathbb{Z}^*_{2p_k}$ – Tim Ratigan Nov 6 '13 at 22:49