Prove that in any group $G$, we have $[G,G]\cap Z(G)\subseteq \operatorname{Frat}(G)$

Question

Let $G$ be a group, we want to prove $[G,G]\cap Z(G)\subseteq \operatorname{Frat}(G)$.

Can you please give some idea how to solve this?

Here $\operatorname{Frat}(G)$ is the Frattini subgroup, the intersection of all maximal subgroups in $G$.

$[G,G]$ is the commutator subgroup of $G$ and $Z(G)$ the center of $G$.

Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? — Hippalectryon, Mar 05 '15 at 16:47
Can you please define the notation $[G,G]$, $Z(G)$ and Frat$(G)$? Thank you. — Gregory Grant, Mar 05 '15 at 16:48
Frat(G)= frattini subgroup is the intersection of all maximal subgroup of G — user220373, Mar 05 '15 at 16:54

Nicky Hekster · Accepted Answer · 2015-03-05T20:33:05.870

5

Let $M$ be a maximal subgroup of $G$. Then either the subgroup $MZ(G)=M$ or $MZ(G)=G$. The first case is equivalent to $Z(G) \subseteq M$. The second case implies that $G'=M'$. So in any case $G' \cap Z(G) \subseteq M$. Whence $G' \cap Z(G) \subseteq Frat(G)$.

Note - there is a generalization for the upper en lower central central series: $\gamma_{n+1}(G) \cap \zeta_n(G) \subseteq Frat(G)$ for any $n \geq 1$. Going a step further, if $\mathcal{V}$ is a variety of groups then the intersection the verbal and marginal subgroups is a subgroup of the Frattini subgroup: $V^*(G) \cap V(G) \subseteq Frat(G)$.

edited Mar 05 '15 at 20:33

answered Mar 05 '15 at 20:26

Nicky Hekster

52,147

@NichyHekster thank you so much and God bless you – user220373 Mar 08 '15 at 10:08
Thanks for the blessing! If you deem my answer as the right one, please tick it as such. – Nicky Hekster Mar 08 '15 at 13:34

Prove that in any group $G$, we have $[G,G]\cap Z(G)\subseteq \operatorname{Frat}(G)$

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