If G is a group of order n, does Lagrange's theorem also imply that there are subgroups of orders dividing n? That is if I have a group of order 15, are there always subgroups of order 1, 3, and 5?
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I think you'll always be able to find a subgroup of order $1$ ;) – Tyler Mar 16 '12 at 02:31
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2What you are asking about is the converse of Lagrange's Theorem. It is known to be false. See this previous question on partial converses, this one on proving the converse is false, and this one about groups for which the converse does hold. – Arturo Magidin Mar 16 '12 at 03:49
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For cyclic groups the converse does hold. In general however, it does not. – sxd Mar 16 '12 at 09:44
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The simplest counterexample is that $A_4$, the group of even permutations on 4 items, has 12 elements, but contains no subgroups of order 6.
I believe the general question of when an $mn$-group contains a subgroup of order $n$ is still an open research area. Cauchy's theorem guarantees that if $p$ is prime, then a group of order $pn$ contains an element of order $p$, and therefore a cyclic subgroup of order $p$. So for your example, every group of order 15 does contain subgroups of orders 3 and 5. (And, obviously, 1.) As N.S. noted, the Sylow theorems are important here.
This page discusses the issue in more detail, including some other circumstances when the converse of Lagrange's theorem holds. I think the Wikipedia article on Lagrange's theorem also discusses the converse.
MJD
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No. It follows from Lagrange Theorem that $G$ has subgroups of order either 3 or 5, but you can't say which....
But in your case, Cauchy Theorem is what you need (or Sylow's theorems).
N. S.
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4It does not follow from Lagrange's theorem that $G$ has subgroups of order either $3$ or $5$. – Qiaochu Yuan Mar 22 '12 at 02:32
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@QiaochuYuan Why not? By Lagrange, an element of $G$ other than the identity has order $3$, $5$, or $15$. If $15$, then $G$ is cyclic, and there are subgroups of orders both $3$ and $15$. Else, etc. What am I missing? – Andrés E. Caicedo Mar 22 '12 at 04:44
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@Andres: perhaps I should have said it does not follow directly from Lagrange's theorem that $G$ has subgroups of order either $3$ or $5$. What I meant is that the statement of Lagrange's theorem doesn't have a logical form that immediately says anything about possible converses. – Qiaochu Yuan Mar 22 '12 at 04:46
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@QiaochuYuan If $|G|=p_1^{\alpha_1}\cdot...\cdot p_k^{\alpha_k}$, it is an immediate consequence of Lagrange that $G$ has an element of order $p_1$ or $p_2$ or ... or $p_k$. Since this result is a simple Corollary to LT, I would say it follows from LT... – N. S. Mar 22 '12 at 12:26