# A Generating Set and Its Image in the Frattini Quotient

This is a homework question I am having some trouble with.

The claim I'm trying to prove is this: Let $\Phi$ be the Frattini subgroup of a finite group $G$. If $S$ is a subset of $G$, and $\bar{S}$ is the image of $S$ under $G\rightarrow G/\Phi$, then $S$ generates $G$ if and only if $\bar{S}$ generates $G/\Phi$.

I have already shown that $\Phi$ is the set of non-generators of $G$. My inclination is to proceed by cases involving whether or not $S$ is itself contained in a maximal subgroup. Any help would be appreciated.

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## 1 Answer

Obviously if $S$ generates $G$ then $\overline{S}$ generates $G/\Phi$. On the other hand, if $\overline{S}$ generates $G/\Phi$, then $S$ and $\Phi$ together generate $G$. But $\Phi$ is the set of nongenerators, so ...

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I'm still a bit confused about the result. For instance, in the forward direction, I'm trying to show that the collection of equivalence classes given by the image of S cannot be contained in a proper subgroup of the quotient. I understand that this should follow from the fact that S itself has this property in G and that every element of Phi is in a maximal subgroup. Could you perhaps explain in some more detail? Thanks. –  Alexander Sibelius Sep 24 '12 at 5:29
The forward direction (by which I assume you mean "$S$ generates $G$ implies $\overline{S}$ generates $G/\Phi$") is a very general result and has nothing to do with Frattini subgroups. The forward direction is true for any group $G$ and any normal subgroup $\Phi$; in other words, any generating set of any group maps to a generating set of any quotient. –  Ted Sep 24 '12 at 5:57
Ah, I see. Thanks for your help. –  Alexander Sibelius Sep 24 '12 at 6:26