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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?

Shikhar
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    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 a $T$ has index $4$. If $n_{3} =1$, then $G/T$ is a group of order $8$, so it has a subgroup of index $4$, and by the correspondence theorem so has $G$. – Andreas Caranti Feb 23 '22 at 14:26

2 Answers2

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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 $4$.
Andreas Caranti
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  • How can we say normaliser of $T$ is normal in $G$ and the corresponding index is $4$? – Messi Lio Jan 13 '23 at 10:45
  • @MessiLio, the normaliser of a Sylow subgroup is self-normalising, so if it is normal in $G$, it coincides with $G$. If $n_{3} = 4$, then by Sylow's theorems the nornaliser has index $4$, but I am not claiming it is normal in $G$. – Andreas Caranti Jan 13 '23 at 11:19
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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.

Dietrich Burde
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