Question as in title.

Factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24.

My approach is rather bullish - I'm simply trying to find subgroups of order n for each factor.

Subgroup of order 1: $<()>$ (Trivial)
Subgroup of order 2: $<(1 2)>$
Subgroup of order 3: $<(1 2 3)>$
Subgroup of order 4: $<(1 2 3 4)>$
Subgroup of order 6: $S_3$ (Symmetric Group on 3 Points)
Subgroup of order 12: $A_4$ (Alternating Group on 4 Points)

But I cannot deduce the others. Is there a more elegant "solution" to this problem. The question cropped up in an introductory group theory class - so I would really appreciate simple answers - I'm familiar with Lagrange's theorem, cyclic groups etc.

Thanks for any help,


  • 3
    $\begingroup$ You are only missing 8. What's the symmetry group of the square? $\endgroup$ – Angina Seng May 8 '17 at 18:37
  • $\begingroup$ @LordSharktheUnknown Ah thank you - D4. Although is there a more elegant way? $\endgroup$ – Jack May 8 '17 at 18:39
  • 3
    $\begingroup$ This is the sort of question where you just have to hack through all cases. $\endgroup$ – Angina Seng May 8 '17 at 18:40
  • 5
    $\begingroup$ Can the Sylow theorems be applied to prove the existence of a subgroup of order $8$, as $8$ and $4!/8$ have no common factor? $\endgroup$ – avs May 8 '17 at 18:41

It is well-known that $S_4$ is a Lagrangian group, so that the converse of Lagrange's Theorem is true: for each divisor $d$ of $24$ there is a subgroup of $S_4$ or order $d$. Enumerating all subgroups of $S_4$ "by hand" has been done several times at MSE, e.g., see here, with nice and simple answers.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.