0
$\begingroup$

Let G be a finite non-abelian group of order 39. Then find the number of subgroups of order 3 in G? I don't know how to find I. please help

  • 0
    Do you have a presentation of your group ? ... then ask which elements are of order 3. – 2017-02-27

2 Answers 2

2

For any group of order $pq$ with primes $pthis answer. It also shows in the proof how many subgroups there are of order $p=3$, for $(p,q)=(3,13)$, namely $q=13$.

1

Hints: make sure you can add details and explanations to the following

1) There is one unique subgroup of order $\;13\;$ which is thus normal

2) If there is one unique subgroup of order $\;3\;$ then it is normal and thus $\;G\;$ is the direct product of abelian groups

3) Thus, there must be exactly $\;13\;$ groups of order $\;3\;$ in $\;G\;$ .