If Z/pqZ* is the set of all integers g mod pq with gcd(g, pq) = 1, under multiplication, why can't this group be cyclic? I've looked at the order of the group, which, if I'm correct is (p-1)(q-1). This doesn't help me much. I'm supposed to show that the group cannot have an element with that order, but I can't seem to do this. I've looked at a few small groups like Z/15Z*, and the only think I've noticed is that all the elements have even order, but this helps very little as (p-1)(q-1) always will be a multiple of four.
Asked
Active
Viewed 83 times
0
-
It is cyclic if either $p$ or $q$ is $2$ ... – Troposphere Apr 17 '21 at 16:39
-
For another way: By the Chinese remainder there are four solutions of $x^2=1$ in $\Bbb{Z}_{pq}^*$. Namely the four cosets $\overline{a}$ with $a\equiv\pm1\pmod p$ as well as $a\equiv\pm1\pmod q$. All four sign combinations are possible and distinct, when $p,q\ge3$. In a cyclic group you can have at most two solutions to $x^2=1$. – Jyrki Lahtonen Apr 17 '21 at 16:46