# How many subgroups of a 24-group who has 8 elements with order 3?

The question is, for a group $G$, and $|G|=24$. We already know it has exactly $8$ elements whose order is $3$. Then how many subgroups does $G$ have.

Here is what I do:

Since it has $8$ elements whose order is $3$, it has $4$ subgroups whose order is $3$, thus $n_3=4$. It also has $2$ trivial subgroups. And by Sylow's third theorem, it has only one Sylow-subgroup of order $8$. But how can I know the structure of this Sylow-subgroup? And I'm not so sure how to deal with other subgroups. Maybe by director product?

Thank you!

• It's not true that there's necessarily only one Sylow 2-subgroup. – Steve D Sep 5 '17 at 6:23
• @SteveD Sylow 2-subgroup? But its order is 24 – Edward Wang Sep 5 '17 at 6:26
• The symmetric group on 4 elements has order 24, and no normal Sylow subgroups. – Steve D Sep 5 '17 at 6:27
• @SteveD Oh you are right... It should have 3 Sylow 2-subgroups. – Edward Wang Sep 5 '17 at 6:32
• @SteveD I should have said, it's possible that it has 3 Sylow 2-subgroups. – Edward Wang Sep 5 '17 at 6:33

This question does not have a unique answer. There are three isomorphism classes of groups of order $24$ that have $8$ elements of order $3$: $S_4$, $A_4 \times C_2$, and ${\rm SL}(2,5)$.