Timeline for Is a profinite group with a finite number of simple quotients and Jordan-Hölder factors finitely generated?

6 events

when toggle format	what		by	license	comment
Jan 9, 2012 at 21:46	vote	accept	Maurizio Monge
Jan 9, 2012 at 17:59	history	edited	YCor	CC BY-SA 3.0	correction
Jan 9, 2012 at 0:01	comment	added	YCor		If $S$ is any nonabelian group $S^n$, the diagonal embedding of $S$ is not a normal subgroup! Hint: for $n=2$, compute $(g,1)(h,h)(g,1)^{-1}$... In a finite product $\prod_{i\in I}S_i$ of nonabelian simple groups, the only normal subgroups are the $\prod_{i\in J}S_i$ for $J$ ranging over subsets of $I$. It's an easy verification.
Jan 5, 2012 at 19:43	comment	added	Maurizio Monge		I think that I'm missing something... in $S\wr{}G=S^{\|G\|}\rtimes{}G$, isn't the subgroup of $S^{\|G\|}$ formed by vectors with equal components a normal subgroup? In the case of $S$ elementary abelian the normal subgroups are exactly the sub-representations, why in the case that $S$ is non-abelian there should be no invariant subgroup? If $S$ is replaced by $S^{m_n}$, furthermore, all elements of $(S^{m_n})^{\|G\|}$ tuples with the first component equal (of the $m_n$) appears to be invariant under $G$.
Jan 5, 2012 at 0:01	history	edited	Alain Valette	CC BY-SA 3.0	added 2 characters in body
Jan 4, 2012 at 23:55	history	answered	YCor	CC BY-SA 3.0