Every group of order 399 is abelian and not simple

Asked Feb 18 '19 at 13:20

Active Feb 18 '19 at 13:36

Viewed 96 times

Let $G$ a group with order $399$ = ${3*7*19}$, from Sylow-theorem, I know $n_{3}$ must be $1$ or $7$ or $19$ or $133$, for $n_{7}$ must be $1$ or $57$ and for $n_{19}$ must be 1. I know $n_{19}$ is $1$ for every cases, then $G$ is not simple, but I try to study if $G$ is abelian with a derivate subgroup and I can't do this.

How can I study if $G$ is abelian?

edited Feb 18 '19 at 13:36

Andrews

3,953

asked Feb 18 '19 at 13:20

angie duque

2

It is not true that every group of order $399$ is abelian. There is a nonabelian group of order $21$. – Derek Holt Feb 18 '19 at 13:28
"How can I study if $G$ is abelian?" - have a look at this duplicate. Also, the group cannot be simple, see this duplicate. – Dietrich Burde Feb 18 '19 at 13:31
Ohh, i try to find a nonabelian group of order $399$, consider case $n_{19}$ = $1$, $n_{7}$ =$1$ and $n_{3}$ = $7$. let $H$ a Sylow $19$- subgroup and $K$ a Sylow 7 -subgroup, then $HK$ is a normal subgroup of $G$ and $M$ a Sylow 3-subgroup, then i try to do semidirect product of $HK$ and $M$, but is hard. – angie duque Feb 18 '19 at 13:32
You get a nonabelian group as the direct product of a nonabelian group of order $21$ with a cyclic group of order $19$. – Derek Holt Feb 18 '19 at 13:37
I ask if $G$ is abelian. In the otherwise , i try to find a nonabelian group with a semidirect product. :) – angie duque Feb 18 '19 at 13:43
Ohhhh yeah, i have to study direct product of order 21. Thank you – angie duque Feb 18 '19 at 13:45
No, you need a semidirect product of order $21$, and then a direct product. – Jyrki Lahtonen Feb 18 '19 at 17:58

Every group of order 399 is abelian and not simple

0 Answers0