Show that a group $G$ so that the order of the quotient group is $| G/C(G) | = 37$ doesn't exist.

Question

I guess $C(G)$ here means the center of the group, I am not sure because this is an exam question from a previous year. I tried by using Lagrange's theorem but don't know where to go from there. Any hint helps! Thanks.

@LordSharktheUnknown which means what? Sorry I am a begginer in abstract algebra. — user15269, Jun 30 '19 at 11:26
It's a hint: that $G/C(G)$ having prime order is significant. — Angina Seng, Jun 30 '19 at 11:27
G/C(G) is cyclic (of prime order) implies G is abelian. Proofs to be found on MathStackExchange. — Nicky Hekster, Jun 30 '19 at 11:30
If you are a beginner in abstract algebra, start reading this question first. — Dietrich Burde, Jun 30 '19 at 11:57

score 3 · Accepted Answer · edited Jun 30 '19 at 11:34

3

Hint: If $G/Z(G)$ is cyclic, then $G$ needs to be abelian and therefore we get $G/Z(G) = \{\bar{e}\}$. Why is the quotient group cyclic in your case?

edited Jun 30 '19 at 11:34

Nicky Hekster

52,147

answered Jun 30 '19 at 11:28

Con

9,119

Show that a group $G$ so that the order of the quotient group is $| G/C(G) | = 37$ doesn't exist.

1 Answers1