# Is there a convenient way to show that the symmetric group $S_4$ has a subgroup of order $d$ for each $d|24$?

Show that the symmetric group $S_4$ has a subgroup of order $d$ for each $d|24$.

From Lagrange's theorem I know that if $G \le S_4$, then the order of $G$ necessarily divides $|S_4|=24$. However the question actually asks the converse of the Lagrange's theorem, so I cannot apply the theorem directly. (And I don't think I can apply it indirectly, either)

So my question is:

Instead of listing the subgroups in detail, is there a convenient way of proving the existence of subgroups of every $d|24$?

One convenient way could be to consider only composite numbers: $2.3, 2^2.3, 2^3.3$.

For divisor $2.3$, natural subgroup is $S_3$ (without looking the list, we can say, it is a natural candidate for subgroup of this order).

For divisor $2^2.3$ again, natural subgroup $A_4$.

• How about order 8? – Rescy_ Nov 21 '15 at 3:43
• It is order of Sylow-$2$ subgroup. Do you expect it to be obtained in some way? Then there is a geometric way for it. – Groups Nov 21 '15 at 3:48

If $d|24$, $d\in {1,2,3,4,6,8,12,24}$.

$d=1: \{e\}$

$d=2:\{e,(1 2)\}$

$d=3:\langle(1 2 3)\rangle$

$d=4:\langle(1 2 3 4)\rangle$

$d=6: S_3$

$d=8:\langle(1 2), (2 3), (3 4)\rangle$

$d=12: A_4$

$d=24: S_4$