Let p be an odd prime and let m = 2p. Show that there is a primitive root mod m, in the sense that there is a number a between 1 and m − 1 with order φ(m) = p − 1 mod m. i'm not able to figure out to proceed. any help i highly appreciated .

  • $\begingroup$ Do you know that there is a primitive root mod $p$ ? $\endgroup$ – lhf Oct 28 '16 at 14:55


Chinese remainder theorem: $\;\begin{aligned}[t]\mathbf Z/m\mathbf Z&\simeq \mathbf Z/2\mathbf Z\times\mathbf Z/p\mathbf Z\\ x\bmod m&\mapsto(x\bmod 2,x\bmod p)\end{aligned}$

The multiplicative group $(\mathbf Z/p\mathbf Z)^\times$ is cyclic, i.e. it is the set of powers of a single element.

Some details:

Let $2u+vp=1$ a Bézout's relation between $2$ and $p$. If $\alpha$ is a generator of the cyclic group $(\mathbf Z/p\mathbf Z)^\times\;$ represented by a number in $[2\,..\,p-1]$, the element $(1,\alpha)\in(\mathbf Z/2\mathbf Z)^\times\times(\mathbf Z/p\mathbf Z)^\times\;$ has order $p-1$, and it corresponds by the inverse isomorphism to the element $2\alpha u+vp\;$ in $\;\mathbf Z/2p\mathbf Z$.

  • $\begingroup$ can you explain in more detail ? $\endgroup$ – kkr Oct 28 '16 at 12:19
  • $\begingroup$ @kkr: please see my completed answer. Is it clearer? $\endgroup$ – Bernard Oct 28 '16 at 13:01
  • $\begingroup$ can you explain it in more simpler way? I don't know what a cycli group is and what isomorphism is $\endgroup$ – kkr Oct 28 '16 at 13:11
  • $\begingroup$ A cyclic group is a group generated by he powers of a single element. A (ring) isomorphism such as mentioned at the beginning, is a map between rings which respects addition and multiplication in these rings. I've added details of the isomorphism in the Chinese remainder theorem. $\endgroup$ – Bernard Oct 28 '16 at 14:53

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