# How to prove that if n divides 4! there is a subgroup of order n in S4.

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,

Jack

• You are only missing 8. What's the symmetry group of the square? – Angina Seng May 8 '17 at 18:37
• @LordSharktheUnknown Ah thank you - D4. Although is there a more elegant way? – Jack May 8 '17 at 18:39
• This is the sort of question where you just have to hack through all cases. – Angina Seng May 8 '17 at 18:40
• 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? – 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.