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I am trying to get a better feel for what the Frattini subgroup really is, intuitively.

Let $G$ be a group and denote its Frattini subgroup by $\Phi(G)$. I know that $\Phi(G)$ is the intersection of the maximal subgroups of $G$, and I know that it is the set of 'non-generators' (Isaacs calls them 'useless' elements) of $G$, i.e. elements $u$ for which if $\langle X \cup \{u\} \rangle =G$, then $\langle X \rangle = G$, or equivalently, if $\langle X \rangle \ne G$, then $\langle X \cup \{u\} \rangle \ne G$, where $X \subseteq G$ is a subset of $G$, and $u \in \Phi(G)$.

Since $\Phi(G)$ is the set of these elements, it would help to better understand what exactly these elements are. Is it true that such an element $u \in \Phi(G)$ is necessarily a product of elements in $X \subseteq G$ ($u$ and $X$ as above)? If not, what is an example where it isn't?

Finally, where exactly does the connection lie between these 'non-generators' and (the intersection of) maximal subgroups? How do we see that they must lie in a maximal subgroup, and conversely that if an element lies in all maximal subgroups then it must be a 'non-generator'?

Thanks for the help, as always.

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What is $X$ in the third paragraph? –  user641 Mar 14 '13 at 3:17
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Let me plug this related question of mine (and Jack Schmidt's excellent answer). –  Alexander Gruber Mar 14 '13 at 7:28
    
@SteveD: $X$ and $u$ are as they are explained at the end of the second paragraph. –  Alex Petzke Mar 14 '13 at 13:49
    
Let me stress that what you replied does not answer my question. In the second paragraph, $X$ can be any set that generates $G$. So of course any element of $G$ (including the ones in the Frattini subgroup) are a product of elements of $X$. So that question seems to hold no meaning. –  user641 Mar 14 '13 at 23:16
    
@SteveD: Well, then it may not have meaning. It was just my attempt to make some observations on something I knew very little about. –  Alex Petzke Mar 15 '13 at 1:29

2 Answers 2

up vote 7 down vote accepted

The equivalence between the two conditions is purely formal, it holds in every category of algebraic structures.

If $u$ is a non-generator and $H$ is a maximal subgroup of $G$ (by which we mean of course a maximal proper subgroup), then $\langle H \rangle = H \neq G$, hence $\langle H,u \rangle \neq G$, which implies $H = \langle H,u \rangle$ since $H$ is maximal, and therefore $u \in H$. Hence, $u$ lies in every maximal subgroup.

If $u$ is not a non-generator, choose some $X \subseteq G$ with $\langle X \rangle \neq G$ but $\langle X,u \rangle = G$. By Zorn's Lemma there is a subgroup $H$ which is maximal with the property that it contains $X$, but does not contain $u$. In fact, $\langle X \rangle$ is such a subgroup, and if $\cal C$ is a non-empty chain of such subgroups, then one can easily check that $\cup \cal C$ is a subgroup with this property. Observe that $H$ is maximal: If $K$ is a subgroup containing $H$ properly, we must have $u \in K$ and $X \subseteq K$, hence $K=G$. Hence, $H$ is a maximal subgroup not containing $u$.

Remark: Not every subgroup of a group can be enlarged to a maximal subgroup. In fact, there are groups (such as $\mathbb{Q}$) with no maximal subgroups at all. Therefore the proof is somewhat clumsy, but it works.

More generally, if $G$ is any algebraic structure, then the intersection of all proper substructures of $G$ is called the radical of $G$, and by the proof above it coincides with the set of all non-generators of $G$. If $G$ is a group, we get the Frattini subgroup. If $G$ is a left module over a ring $R$, we get its radical, which in the particular case of $G=R$ is known as the Jacobson radical. So the Frattini subgroup is really just a special case of a more general construction, whose special cases one might be familiar with.

Probably the best way to get familiar with the Frattini subgroup is to learn some of its nice properties. It is always a characteristic subgroup. If $G$ is a finite group, then $\Phi(G)$ is nilpotent. If $G$ is a finite $p$-group, then $\Phi(G)$ is the smallest normal subgroup whose quotient is elementary abelian. In this situation, Burnside's Basis Theorem states that a subset generates $G$ if and only if its image generates the $\mathbb{F}_p$-vector space $G/\Phi(G)$, which reduces the former condition to linear algebra.

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+1 for the vector space interpretation of $G/\Phi(G)$. To add to that: we then ascertain that the frattini quotient of a nilpotent group $G$ is the product of elementary abelian groups corresponding to each prime dividing $|G|$. –  Alexander Gruber Mar 14 '13 at 7:36
    
That helps, thanks! –  Alex Petzke Mar 15 '13 at 19:44

Take the cyclic group $\,G:=\langle\,x\,\rangle\,$ of order $\,p^2\,\,\,,\,\,p\,$ a prime. Then $\,\Phi(G)=\langle\,x^p\,\rangle\,$ (it's easy to check this taking $\,G\,$ as a vector space over $\,\Bbb F_p=\Bbb Z/p\Bbb Z\,$ of dimension $\,2\,$).

We know that the generators of $\,G\,$ are the elements $\,x^i\,$ , with $\,(i,p)=1\iff p\nmid i\,$ , and thus all the elements of the form $\,x^{kp}\,\,,\,\,k\in\Bbb Z\,$ , are the ones that cannot generated $\,G\,$, i.e. the non-generators.

Finally, if you know the proof of the relation between the Frattini subgroup and the set of non-generators, there you can see that an element that belongs to all the maximal subgroups of $\,G\,$ has to be a non-generator as otherwise it together with some other subset would generate the whole group without being possible to drop this element from the whole generating set, and from here one can construct a maximal subgroup that won't contain that element (Zorn Lemma's calling in the general, non-finite case)...

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