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I know that $A_{4}$ is a counterexample to converse to Lagrange's Theorem. I was thinking that I don't know another one. What another examples is possible?

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There are lots of divisors of the order $S_6$ that do not correspond to subgroups. I can't remember exactly which ones, but the Sylow Theorems will let you know. –  Joe Johnson 126 Jan 28 '12 at 14:37
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Here are 2 related questions: math.stackexchange.com/q/66451 and math.stackexchange.com/q/100933 –  Jonas Meyer Jan 28 '12 at 15:07
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A stronger version of the converse to Lagrange's theorem requires that if HG and d divides $[G:H]$, then there is an intermediate subgroup K, HKG, such that $[H:K]=d$. CLT just requires this when H = 1. This stronger condition is equivalent to G being supersoluble, and A4 is the smallest non-supersoluble group. In some sense the key problem is having chief factors of composite order, since one has a hard time splitting them up (A4 has a chief factor of order 4, and so the 3 cannot be used with only part of the 4). AGL(1,p^2) in general works, not just AGL(1,4)=A4. –  Jack Schmidt Jan 28 '12 at 17:39
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2 Answers 2

up vote 4 down vote accepted

There is a class of groups which satisfy the converse to Lagrange's theorem; appropriately, they're called CLT groups. That is, $G$ is a CLT group if $|G|=n$ and for each $d|n$ there is a subgroup of $G$ of order $d$.

Every CLT group is soluble, meaning that there is a chain $$ \{ e \} = H_0 \trianglelefteq H_1 \trianglelefteq \cdots \trianglelefteq H_r = G$$ of subgroups for which $H_i/H_{i-1}$ is abelian for each $i$.

So any group which isn't soluble is a counterexample to Lagrange.

For instance, $S_n$ is not soluble for $n \ge 5$, so $S_n$ does not satisfy the converse to Lagrange's theorem whenever $n \ge 5$.

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I would like to elaborate a bit on a detail of Clive's answer. One way to see that CLT-groups are solvable is through Hall's theorem.

Definition. Let $\pi(n)$ denote the set of prime divisors of $n$. For any $\mathcal{S}\subseteq \pi(|G|)$, a Hall $\mathcal{S}$-subgroup of $G$ is a subgroup $H$ with $\text{gcd}(|H|,[G:H])=1$ such that $\pi(|H|)=\mathcal{S}$.

Hall $\mathcal{S}$-subgroups are generalizations of Sylow $p$-subgroups, which are Hall $\{p\}$-subgroups. If the converse to Lagrange was true for all groups, Hall subgroups would always exist. Unfortunately, this is not the case. The following is a well-known extension of a theorem of Hall.

Theorem. Hall subgroups exist for every $\mathcal{S}\subseteq \pi(|G|)$ if and only if $G$ is solvable.

Thus, in particular, for every nonsolvable group $G$, there must be some $\mathcal{S}\subseteq \pi(|G|)$ for which $G$ does not contain a group whose order is the product of the highest powers of each $p\in\mathcal{S}$ which divide $|G|$.

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