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 $A_4$.

  • $\begingroup$ How about order 8? $\endgroup$ – Rescy_ Nov 21 '15 at 3:43
  • $\begingroup$ 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. $\endgroup$ – Groups Nov 21 '15 at 3:48

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

$d=1: \{e\}$

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

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

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

$d=6: S_3$

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

$d=12: A_4$

$d=24: S_4$


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.