Group of order 24 has subgroup of index 4 I have a question .
A group of order 24 will have a subgroup of index 4 .
Is this always true ?
I tried proving but this can't find any way  .
I searched for classification of group of order 24  on Wikipedia.
It is given  there are 15 non- isomorphic groups of order 24 and each has a subgroup of order 6 .
So here definitely I can conclude that there is subgroup of index 4 in every group of order 24.
Now , I want to show this explicitly.
I tried to show it using class equation by considering possibilities of cardinality of conjugacy class , but i couldn't show that class equation necessarily has a conjugacy class of Cardinality 4 .
Can someone please help me with this ? What are possible approach to tackle such problem?
 A: I have slightly expanded my comment above into an answer.
Let $G$ be a group of order $24$, let $T$ be a Sylow $3$-subgroup, and let $n_{3}$ be the number of Sylow $3$-subgroups.
Sylow's theorems imply that $n_{3}$ is $1$ or $4$.

*

*If $n_{3} = 4$, then the normaliser of $T$ has index $4$.

*If $n_{3} =1$, then $T$ is normal in $G$, and $G/T$ is a group of order $8$, so it has a subgroup of index $4$, and by the correspondence theorem so has $G$. Or, in a more elementary fashion, if $D$ is a subgroup of $G$ of order $2$ (which exists by Cauchy's theorem), then $T D$ is a subgroup of index $4$.

A: Andrea's answer is best possible. However, in case you are familiar with supersolvable groups you can check your list you have - for any finite supersolvable group, there are subgroups of every possible order, i.e., for $|G|=24$ there are proper nontrivial subgroups of orders $2,3,4,6,8,12$, hence of index $12,8,6,4,3,2$. This is "Langrange Converse", see for example the post
Complete classification of the groups for which converse of Lagrange's Theorem holds
Now in your list, from the $15$ groups, all but three are supersolvable, namely $S_4$, $A_4\times C_2$ and $SL(2,3)$ are not supersolvable. For them, it is true by direct inspection.
