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

Question

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?

Answer 1

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".

Answer 2

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.