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I know converse of Lagranges theorem is not true in general. I find example on $A_{n}$ and $S_{n}$.

My question is : can we find an example of a group other than $S_{n}$ and $A_{n}$ where converse of Lagranges theorem is not true?

Pradip
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  • Is $A_4 \oplus \Bbb Z_3$ allowed? – Chinnapparaj R Jun 30 '19 at 03:09
  • @Chinnapparaj R Yes, allowed... Does the group serve my purpose? – Pradip Jun 30 '19 at 03:45
  • There are examples. Take a look at https://math.stackexchange.com/a/100934/403337. See for instance the paper on "non CLT groups". You can tell from their orders that some of these groups are not $S_n$ or $A_n$. Other than the $24$ examples with order less than $100$ claimed to be exhibited in the paper, they all are "supersolvable". –  Jun 30 '19 at 04:02

2 Answers2

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Consider $G=H/Z(H)$ where $H=SL(2,\Bbb Z_7)$.This group is a simple group of order $168$. So this group has no subgroup of order $84$, even though $84$ divides $168$.

Chinnapparaj R
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One counterexample is given by $G = \text{SL}_2(\mathbb{F}_3)$ as $G$ has no subgroup $H \subset G$ of order $12$. That is because the quotient would have order $2$ and therefore is an abelian group such that we get a surjective homomorphism $G^{\text{ab}} \rightarrow H$. Since the order of $G^{\text{ab}}$ is $3$, that cannot be the case. Note that this depends very much on the cardinality of the field $\mathbb{F}_3$, as the group $\text{SL}_n(K)$ is perfect except for $n = 2$ and $K$ of cardinality $2$ or $3$.

Oops, did not see that you excluded $S_n$ as well. Just ignore this: Another counterexample is the group $S_n$ for $n \geq 5$ as $A_n$ is simple then. Therefore $S_n$ cannot have a subgroup of order $\frac{n!}{4}$.

Let me state a way to find some counterexamples:

Let $G$ be a finite group, $N \subset G$ a normal subgroup and $g \in G$. Then $\text{ord}(\overline{g})$ is a divisor of $\text{ord}(g)$, where $\overline{g}$ is the image of $g$ under the quotient map $G \rightarrow G/N$. If we now assume that $[G:N] = 2$ and that $\text{ord}(g)$ has odd order, then we get $\text{ord}(\overline{g}) = 1$ by Lagrange's theorem, such that $g \in N$. We just showed that every element of odd order needs to be in $N$. Therefore, if more than half of the elements of a group $G$ have odd order, the group cannot have a subgroup of index $2$. That for example holds for $A_4$ since $A_4$ has $9$ elements of odd order. So this is one way of proving that $A_4$ has no subgroup of order $6$.

There is even a characterization for the case of index $2$ subgroups:

Proposition:

A group $G$ has no index $2$ subgroup iff $G = \langle \lbrace g^2 \mid g \in G\rbrace \rangle$.

Con
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  • What is G$^{ab}$? And How can you get such surjective homomorphism? – Pradip Jun 30 '19 at 19:13
  • That is the abelianization of $G$ defined as the quotient $G/[G,G]$. It is the biggest abelian quotient as one exactly killed the commutators. Therefore one gets a morphism into every abelian quotient by projecting. One can also state all this as a universal property. Whenever you have a morphism $f \colon G \rightarrow H$ into an abelian group there is exactly one morphism $g \colon G^{\text{ab}} \rightarrow H$, such that $g \circ \eta = f$, where $\eta$ is the quotient map from $G$ to $G^{\text{ab}}$. – Con Jun 30 '19 at 19:19