Talk:Frattini subgroup
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example : p2
editWhy is the example given as p2? Why not just p? Or n? Any cyclic group has Φ(G)=G, so why the fancy-ness? I know it's petty, but it could also confuse people... 88.109.188.112 08:47, 10 May 2007 (UTC)
No finite or cyclic group has Phi(G)=G. The smallest group with nontrivial Frattini subgroup is the cyclic group of order four. It has a unique maximal subgroup, a cyclic subgroup of order two, which is therefore its Frattini subgroup. JackSchmidt 19:13, 6 June 2007 (UTC)
- Well, which is it? "Any" cyclic group has Φ(G)=G, or "no" cyclic group has Phi(G)=G? Or is Φ(G) not the same as Phi(G)? Or maybe ... well, who knows ... Timothy Perper 17:37, 1 December 2007 (UTC)
- "No". The first commenter 88.109.188.112 was confused. JackSchmidt 18:56, 1 December 2007 (UTC)
- Actually, a number of cyclic groups have Φ(G)=G, namely those cyclic groups of prime order. The reason is in the "if G has no proper maximal subgroups, then Φ(G) is defined to be G itself" line (that is not always included in the definition, but on this page is) I'll add something in the text to clear up the confusion...SetaLyas (talk) 21:17, 14 December 2008 (UTC)
- No. The identity subgroup is a maximal subgroup of a cyclic group of prime order, and the Frattini subgroup of such a group is the identity subgroup. The Frattini subgroup of a finite p-group is the smallest normal subgroup such that the quotient is an elementary abelian p-group. A cyclic group of prime order is already an elementary abelian p-group. No finite non-identity group is equal to its Frattini subgroup. An example of a group equal to its Frattini subgroup is the additive group of rational numbers. JackSchmidt (talk) 06:12, 20 December 2008 (UTC)
- I'm not saying you're wrong JackSchmidt, I'm saying it depends on what definition of the Frattini subgroup you use. I've seen it defined both ways before SetaLyas (talk) 02:22, 15 January 2009 (UTC)
- No. The identity subgroup is a maximal subgroup of a cyclic group of prime order, and the Frattini subgroup of such a group is the identity subgroup. The Frattini subgroup of a finite p-group is the smallest normal subgroup such that the quotient is an elementary abelian p-group. A cyclic group of prime order is already an elementary abelian p-group. No finite non-identity group is equal to its Frattini subgroup. An example of a group equal to its Frattini subgroup is the additive group of rational numbers. JackSchmidt (talk) 06:12, 20 December 2008 (UTC)
- Actually, a number of cyclic groups have Φ(G)=G, namely those cyclic groups of prime order. The reason is in the "if G has no proper maximal subgroups, then Φ(G) is defined to be G itself" line (that is not always included in the definition, but on this page is) I'll add something in the text to clear up the confusion...SetaLyas (talk) 21:17, 14 December 2008 (UTC)
- "No". The first commenter 88.109.188.112 was confused. JackSchmidt 18:56, 1 December 2007 (UTC)
Contradiction?
edit- So if G has no (proper) maximal subgroups, then Φ(G) is G itself.
- (I removed the "proper", because a maximal subgroup is always proper, according to the article.)
- Φ(G) is equal to the set of all non-generating elements of G.
But how could G itself be the set of all non-generating elements? Which elements would generate the group, when all elements are non-generating? To me that makes only sense, when G is the trivial group. But then, I think, it shouldn't sound, as if there could be other cases, where G has no maximal subgroups. By the way: In Maximal subgroup the Frattini subgroup is defined as the intersection of the m.s. - not of the m.s. and G.
To me the following start would make some sense:
- In mathematics, the Frattini subgroup Φ(G) of a group G is the intersection of all maximal subgroups of G. For the case that G is the trivial group 0, which has no maximal subgroups, it's defined as Φ(0) = 0.
Zorn's lemma
editThe proof that the intersection of all maximal subgroups is equal to the set of non-generators uses Zorn's lemma, a form of the axiom of choice. One inclusion does not need Zorn's lemma or the axiom of choice: if x is a non-generator and M is a maximal subgroup of G, then since M is maximal, the subgroup generated by M and x is either M, which means that x is in M, or G, which cannot be the case because that together with the fact that x is a non-generator imply that M, which is already a subgroup of G, is equal to G, a contradiction. Therefore, x lies in every maximal subgroup of G. GeoffreyT2000 (talk) 15:10, 8 October 2015 (UTC)