Assume $N \triangleleft G$ Prove that if $[G:H]$ is a prime, then $G/N$ is cyclic. prove or disprove the converse of the first statement.

Question

Assume $N \triangleleft G$ Prove that if $[G:N]$ is a prime, then $G/N$ is cyclic. prove or disprove the converse of the first statement.

Doesn't LaGrange state that if the group is prime, then it is cylic? Does that still stand for this?

Answer 1

Doesn't LaGrange state that if the group is prime, then it is cylic?

No, it does not. Lagrange's theorem states that $|H|$ divides $|G|$ for any group $G$ and any subgroup $H$. It is a consequence that if $|G|$ is prime then $G$ is cyclic (by considering any non-trivial cyclic subgroup).

Assume $N \triangleleft G$ Prove that if $[G:H]$ is a prime, then $G/N$ is cyclic.

I assume that you actually meant "$[G:N]$ is prime". Then yes: this follows from Lagrange's theorem. The $N$ subgroup is just a noise. Because in this scenario Lagrange applies to $G/N$ directly, regardless of what $N$ is.

Answer 2

[G:N] is the number of cosets of G, which are the elements of $G/N$ and hence the order of $G/N$. Then if [G:N] is prime, we have that $G/N$ must be cyclic since:

For any cyclic group H, where the order of H is some prime p , we have every for every h $\in$ P where h $\neq$ 1 that $|<h>|$ divides P, hence $|<h>|$ = 1 or $p$, but if the order is 1, this implies $h$ = 1, hence $|<h>|$ = $p$. Hence the cyclic subgroup generated by $h$ generates all of H. Hence H is cyclic.

Since H was any arbitrary group, this argument applies to $G/N$ as well.

The converse statement would be if $G/N$ is cyclic then the index of N is prime where N is a normal subgroup. This is not true in general as:

Consider the group H = $<(1234)>$ as a cyclic subgroup of S$_4$. H is a group, and 1 $\in$ H. ${1}$ is a normal subgroup of H and $H/1$ = $H$. ${1}$ has index 24, which is not prime. The index is 24 since the order of S$_4$ is 4*3*2*1 = 24.

However a different result holds in that if $G/N$ is cyclic, then $G$ is abelian.