Let $G$ be a finite group, then its Frattini subgroup $\Phi(G)$ is the intersection of all maximal subgroups of $G$. If $N \unlhd G$, then in general we have $$ \Phi(G)N / N \le \phi(G / N) $$ as could be seen easily by noting that the maximal subgroups in $G / N$ correspond to maximal subgroups of $G$ which contain $N$, i.e. they are a subset of the maximal subgroups of $G$.

For $p$-groups, the Frattini subgroup is characterised as the smallest normal subgroup such that its quotient is elementary abelian. Using this, for $p$-groups we have $$ \Phi(G)N / N = \Phi(G / N). $$ As $G / \Phi(G)N$ is, as a homomorphic image of the elemantary abelian group $G / \Phi(G)$, itself elemenary abelian (and nontrivial if $N \ne G$) and $$ G / \Phi(G)N \cong (G / N) / (\Phi(G)N / N) $$ we have for the $p$-group $G / N$ that $\Phi(G / N) \le \Phi(G)N/N$.

Is this right? I nowhere found this result, so I am asking for confirmation.


1 Answer 1


After seeing $\Phi(G)N / N \le \Phi(G / N)$,

$G/\Phi(G)$ is elamantary abelian $\implies$ $G/\Phi(G)N$ elamantary abelian $\implies (G/N)/(\Phi(G)N/N) $ is elemantary abelian. Hence,

$$\Phi(G)N / N \ge \Phi(G / N)$$. Yes, it is true and you can see this in many $p$- groups book.

Another way to see this;

$$\Phi(G)=G'G^p$$ for $p$ groups where $G^p=<\{x^p|x\in G \}>$. As $ G'/N=(G/N)'$ and $G^p/N=(G/N)^p$ the result follows.


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