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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!

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

Their numbers of subgroups are respectively 30, 26, and 15. (I did that by computer - it would be tedious and not particularly instructive to do it by hand.)

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