# Basis of a group

Let $G$ be a finite group. I will say that a set of a subgroups $H_1,\ldots ,H_k$ defines a basis for a group $G$ if any subgroup $H$ of $G$ there exists $S\subset [k]$ such that $H=\cap_{i\in S}H_i$.

My question is does it possible to give any upper bounds on the size of the minimal basis for $G$. For example does it possible to prove that for any $G$ there exists a basis of size at most $|G|^{10}$?

For example for an Abalian group it is easy to show that there is always exists a basis of size $|G|$. For example in case $A=Z_p^n$ it will be all maximal subgroups.

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On the abelian case: The maximal subgroups do not form a basis. Think of a cyclic group of prime power order, where the subgroups form a chain and hence you need all of them in a basis (though I think your upper bound would still hold). –  Florian Eisele Aug 14 '12 at 9:56
Evidently your basis must contain all maximal subgroups. A little googling finds that Liebeck, Praeger and Saxl have a result describing almost all the maximal subgroups of the symmetric and alternating groups. The description is rather complicated, but you could probably use it to get a lower bound. –  HJRW Aug 14 '12 at 9:57
To Florian Eisele: thanks for a comment I made a change. –  Klim Efremenko Aug 14 '12 at 10:11
The name basis of a group I invented just now, may be this notion have an other name in the literature. –  Klim Efremenko Aug 14 '12 at 10:15
Why the tags "lie-groups" and "rt.representation-theory"? –  Frieder Ladisch Aug 14 '12 at 12:40
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Thanks for your comment. In fact I even do not know how to prove that the number of maximal subgroups is at most $|G|$ –  Klim Efremenko Aug 14 '12 at 22:09