Let G be a finite group a) Prove there exists a normal subgroup N of G with the following property: There exists no normal subgroup N' of G with \(N\subset N'\subset G\) b) For a normal subgroup N of G in part a), prove that G/N as a simple quotient of G Please, help.
Is this what we're talking about? http://en.wikipedia.org/wiki/Sylow_theorems
not really, it is about normal
So just normal theorems.
What kind of math is this?
Not that easy. Show me steps, please
don't even what a normal subgroup is lol
I got part a) this is my trying: Assume G is not simple nor abelian Suppose there exist 2 normal subgroups of G, namely N, N', such that N<N'<G N <N', then |N|<|N'| \(N\triangleleft G, hence, |G|=[G:N]~|N|\) \(N'\triangleleft G, hence, |G|=[G:N']~|N'|\) then \([G:N] |N|=[G:N']|N'|\) to get it true, we need [G:N] > [G:N'] which is impossible since N <N' then, our hypothesis is wrong, then there is no 2 normal subgroups of G , N, N' satisfy N < N' . I still need help on part b Appreciate any tips
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