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This is a multiple choice for finite groups. For which one of the following groups, the converse of Lagrange's Theorem is not generally satisfied? I know the converse is true for cyclic groups.

1) All abelian groups

2) All groups of order 8

3) The group $S_4$

4) All groups of order 12

Thank you in advance!

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What did you do so far? How would you formulate that inverse for example? – Hagen von Eitzen Dec 28 '12 at 15:26
I was asked to probe if n didvides |G| then G has a subgroup of that order. I know this is true for cyclic subgroups. – Nancy Rutkowskie Dec 28 '12 at 15:31
up vote 3 down vote accepted

Lagrange states that the order of a subgroup is always a divisor of the original group order. So we can ask if the inverse holds: Given a group $G$ of finite order $n$ and a divisor $d$ of $n$, does there exist a subgroup $H\le G$ of order $d$?

You already pointed out that this inverse is true for finite cyclic groups. Therefore, it also holds for finite abelian groups (item 1): According to the classification of finite abelian groups, these are direct sums of cyclic groups and the product of their orders is $n$. It is possible to obtain $d$ by taking appropriate divisors of the respective subgroups.

ad 2: Let $G$ be of order $8$. Let $1\ne g\in G$. If $\langle g\rangle = G$, then $G$ is cyclic and we are done. If $\langle g\rangle$ is of order $4$, it has a subgroup of order $2$ (and of course of order $1$) and we are done. Therefore, we may assume that $\langle g\rangle$ is of order $2$ for all nontrivial $g$. But a group with $g^2=1$ for all $g\in G$ is abelian (a much-loved beginner's exercise). Therefore the converse also holds for groups of order $8$.

I see that meanwhile complete answers for 3 and 4 are already available, so I'll stop here. :)

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Lagrange states that for a finite subgroup $H$ of a group $G$: $|H|$ divides $|G|$

Converse of Lagrange: Given a group $G$ of finite order such that $|G| = n$, then for every divisor $d$ of $n$, there exists a subgroup $H\le G$ of order $d$.

1) Fundamental Theorem of finitely generated abelian groups: Any abelian group of order $n$ is isomorphic to direct product of cyclic groups (with the product of the orders of those cyclic groups = $n$.) It is possible to obtain subgroups of order $d$ (divisors of $n$) by taking appropriate divisors of the respective subgroups. You already know that the converse of Lagrange's Theorem is true for finite cyclic groups. As for infinite cyclic groups (all isomorphic to $\mathbb{Z}$), the only subgroup of finite order is the identity $\{0\}$ with order $1$, which divides all $n \in \mathbb{Z}$, though one cannot really divide the order of an infinite set. At any rate, Lagrange really only applies to finite groups of order $n$, where $n$ is fixed. It follows that the *converse holds for any [finite] abelian group.

(2) Suppose $G$ is of order $8$. Let $1\ne g\in G$. If $\langle g\rangle = G$, then $G$ is cyclic and we are done. Else, if the group has an element of order $4$, it has an element of order $2$ and we are done. (There are only two groups, up to isomorphism, of order $4$, and both have subgroups of order $2$.) Otherwise, we are left with a group of order $8$ such that each element of $G$ has order of at most 2, which means that the group is abelian...(why?). So we can conclude the converse holds for every group of order $8$.

(3) See this link. What are all the factors of $|S_4| = 24$? You'll see there exists an $H_i\le S_4$ such that for each factor $k_i$ of 24, there is a subgroup $H_i$ such that $|H_i| = k_i$. Check-mark: Converse of Lagrange holds for $S_4$.

(4) $A_4 \le S_4$ has order $12$. But: $A_4$ has no subgroup of order $6$. Why not? Since $6 \mid 12$, and there is no $H\le A_4$ such that $|H| = 6$, the converse of Lagrange fails in the case of $A_4$.

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1) Any abelian Group is a product of cyclic groups. Can you extend the result about cyclic groups?

2) If the group has an element of order 4 you are done, otherwise any element has order at most 2 which implies that the group is abelian.

3) You know the group, and the divisors of $24$. For each of them identify a subgroup:

Subgroups of order $2,3,4$ are generated by a permutation of that order. For $6$, just look at $S_3$ inside. And for $12$ look at $A_4$...

So the only one left is the subgroup of order $8$, which is a $2$Sylow subgroup ;)

4) Can $A_4$ have a subgroup of index 2?

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Hint: I think 1 and 2 is not correct because abelian groups and $p$-groups are all nilpotent.

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Very short Hint... – Nancy Rutkowskie Dec 30 '12 at 14:08

Number 4 is wrong.

A counterexample:

$A_4$ has order 12. Does it have subgroup of order 6? The answer is no it doesn't. A detailed proof can be provided if needed.

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Thanks. What about the 3, if you think Babk's answer is correct. – Nancy Rutkowskie Dec 28 '12 at 15:40
An answer has been given for $S_4$ I agree with N.S. I just want to add that a group of order 8 also exists from Sylow Theorem – epsilon Dec 28 '12 at 15:56

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