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September 12th, 2017, 11:51 AM  #11 
Member Joined: Jan 2015 From: usa Posts: 75 Thanks: 0 
I have another question that I can't answer; please help me if you can: let $G$ a group. I want to show that $\phi(G/\phi(G))=\left\{e\phi(G)\right\}$ where $\phi(G)$ is the Frattini subgroup (i.e. the intersection of all maximal subgroups of $G$). Last edited by skipjack; September 13th, 2017 at 02:52 AM. 
September 12th, 2017, 12:49 PM  #12 
Member Joined: May 2017 From: Russia Posts: 33 Thanks: 4 
What does $\displaystyle \left\{e\phi(G)\right\}$ mean?

September 12th, 2017, 01:23 PM  #13 
Member Joined: Jan 2015 From: usa Posts: 75 Thanks: 0 
$e$ is the trivial element of $G$

September 12th, 2017, 05:03 PM  #14 
Member Joined: Jan 2016 From: Athens, OH Posts: 58 Thanks: 34 
First, you should start a new thread when you have a new question. However: For any group G and normal subgroup N of G, the subgroups of G/N are precisely of the form H/N where H is a subgroup of G that contains N. From here, it should be obvious that if $N\subseteq \phi(G)$, then $\phi(G/N)=\phi(G)/N$. In particular, when N is the Frattini subgroup, the Fratinni subgroup of G/N is trivial. 
September 12th, 2017, 11:16 PM  #15 
Member Joined: Jan 2015 From: usa Posts: 75 Thanks: 0 
And how does that answer the question? You showed that $\phi(G/N)=\phi(G)/N$ how can we conclude that $\phi(G/\phi(G))=\left\{e\phi(G)\right\}$? Thanks. Last edited by skipjack; September 13th, 2017 at 02:50 AM. 

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action, frobenius, group 
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