Talk:Frattini subgroup

example : p²

Why 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)

Contradiction?

• 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.

Lipedia (talk) 07:51, 13 February 2011 (UTC)

Zorn's lemma

The 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)