# Prove that $\mathbb{Z}_{2p}^\ast$ has a generator

I'm working on a problem from a past exam paper,

Prove that if $$p>2$$ is a prime, then $$\mathbb{Z}_{2p}^\ast$$ has a generator.

I'm using the exam paper to study 'backwards', without course notes, for a course starting this September, so much of the material is unfamiliar.

Based on Wikipedia, I'm tempted to say, \begin{align} & \mathbb{Z}_{2p}^\ast \cong \mathbb{Z}_2^\ast \times \mathbb{Z}_p^\ast \cong \mathbb{Z}_p^\ast \\ \therefore \; & \mathbb{Z}_{2p}^\ast \text{ has a generator, as required.} \end{align} but this seems like too little for 7/100 marks. Any tips?

• Do you reckon you're allowed to state without proof that ${\bf Z}_p^*$ is cyclic? Commented Jul 3, 2022 at 3:18
• @GerryMyerson That could be it. Thanks.
– mjc
Commented Jul 3, 2022 at 4:13

Note that if $$n$$ and $$m$$ are relatively prime, then \begin{align} &\mathbb{Z}_{mn}= \mathbb{Z}_n \times \mathbb{Z}_m \\ \Longrightarrow \ \ \ & \mathbb{Z}_{mn}^*= (\mathbb{Z}_n \times \mathbb{Z}_m)^* \\ & \ \ \ \ \ \ \ = (\mathbb{Z}_n^* \times \mathbb{Z}_m^*) \end{align}

Since $$p>2$$ is a prime, following your temptation is correct.

For the first part, it has been explained to me by category theory, let's say, buffs, that for certain reasons, the group of units functor respects products.

That is, if $$\mathcal R_1,\,\mathcal R_2$$ are rings, we get $$(\mathcal R_1×\mathcal R_2)^×=\mathcal R_1^××\mathcal R_2^×$$.

Of course, by CRT, we do indeed have $$\Bbb Z_{2p}\cong\Bbb Z_2×\Bbb Z_p$$.

Finally, as far as $$\Bbb Z_p^×$$ being cyclic, well, we need to find a primitive. I don't (once again) know how off the top of my head.

There's a good proof in Vinogradov's Elements of Number Theory, published I know by Dover.

It certainly became a mantra long ago that "the multiplicative group of a finite field is cyclic".

• Showing ${\bf Z}^{\times)_p$ is cyclic is, at the level of an introductory course in Number Theory, not all that easy. Not as hard as, say, Quadratic Reciprocity, but not all that easy. It has probably been proved here on math.se several times. Commented Jul 3, 2022 at 7:38
• Yes. @GerryMyerson I just read a proof in my Vinogradov. It's quite clever. Starts on the bottom of page $106$ of the latest Dover edition. About a third of a page in length. Commented Jul 3, 2022 at 7:51
• As can happen, it seems that the more general "multiplicative group of a finite field is cyclic" may be easier. There is a neat proof using Euler's formula: $\sum_{d|n}\varphi (d)=n$. @GerryMyerson Commented Jul 6, 2022 at 3:03