Let G be a group of order $n=21$. Now is it possible for converse of Lagrange's theorem to hold? Or if $m \mid n$ will there certainly be a subgroup of order $m$ and if yes are the subgroup unique?
Does converse of Lagrange's theorem hold for a group of order 21?
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group-theory
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1Think about Sylow subgroups. That shuold give the answer. – 2017-01-11
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2If n = 21= 3×7. Then by Sylow's theory of p groups, subgroups of order 3 and 7 exists. But will the subgroups be unique? @астон вілла олоф мэллбэрг – 2017-01-11
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0Have you read the second and third parts of Sylow's theorem? They will be of help here. – 2017-01-11
2 Answers
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The question of @shadow kh refers to the class of finite groups known as CLT groups, where CLT stands for Converse Lagrange Theorem:
$G$ is a CLT group if for each positive integer $d$ dividing $|G|$, $G$ has at least one subgroup of order $d$.
These groups have been studied extensively. It turns out for example that all supersolvable groups are CLT, and all CLT groups are solvable. See also one of the early papers of Henry G. Bray, Pac. J. Math 27 (1968). All $p$-groups, or in general nilpotent groups, are supersolvable. For groups of order $21$ you do not necessarily need Sylow theory, the Theorem of Cauchy suffices: there are elements of order $3$ and $7$ hence subgroups of that order.
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It does not hold in general that if $|G|=n$ and $m|n$ that there is a subgroup of $G$ of order $m$. For example $A_5$ has order $60$ and no subgroup of order $20$.
It may interest you however to look Hall's Theorem or Sylow's Theorem which prove existence of subgroup of order $m$ for some values of $m|n$. As mentioned in comments, for $n=21$ Sylow's Theorem gives the result you're looking for.