Let G be a nontrivial finite group. Prove that there exists subgroups G0, G1,......Gn of G satisfying the following properties: a) $=G_0\subset G_1\subset ....\subset G_n =G$ I have it done. b)$G_{i-1} ~is ~normal ~in~ G_i$, $ 1\leq i\leq n$ c) $G_i/G_{i-1}$is simple for all $1\leq i\leq n$ Please, help. I made a lot of searching but it comes out that is the definition of normal series of a group G. How to prove the definition ??

Question

Let G be a nontrivial finite group. Prove that there exists subgroups G0, G1,......Gn of G satisfying the following properties:
a) $<e>=G_0\subset G_1\subset ....\subset G_n =G$ I have it done.
b)$G_{i-1} ~is ~normal ~in~ G_i$, $ 1\leq i\leq n$
c) $G_i/G_{i-1}$is simple for all $1\leq i\leq n$
Please, help. I made a lot of searching but it comes out that is the definition of normal series of a group G. How to prove the definition ??

loser66 · Answer

Hello, Miracrown, Please, help me in part b, c
Do you want to see my trying in part a?

miracrown · Answer

Alright, go on ...

loser66 · Answer

Let H is a subgroup of G, then G/H ={aH | a in G}
then $a_i H$ is distinct
we can construct:
$\{e\}< H <H \cup a_1H <H\cup a_1H\cup a_2H<.....< G/H$
now label them as {e}= $G_0$
                              H=$G_1$
                  ................................
we have the required series
then, there exists a series like that

loser66 · Answer

for part b) I got stuck at the lemma:
if N < H < G, and N is normal subgroup, then H need not be normal, so that how can I get H is  normal of G?

miracrown · Answer

Errrrm honestly, it's been sometime since I took an algebra class and this is obviously well beyond the level of calculus, so I don't think I can do much to help in a reasonable amount of time. I can still throw out a few ideas if you want, but I can't make any promises.

loser66 · Answer

But if G is abelian, then  we get part b.

miracrown · Answer

Right, trivially so.

loser66 · Answer

Appreciate ANY tips, please

miracrown · Answer

But a majority of algebraically interesting groups are nonabelian.

loser66 · Answer

if it is nonabelian, then to get part b, it must be simple, that is not the case when the given condition is G is a nontrivial finite group.

miracrown · Answer

I'm trying to make sense of your construction of the group sequence.

You're just taking an arbitrary subgroup H<G and then adjoining new cosets incrementally?

loser66 · Answer

yes, because G has many kind of subgroup, some of them are normal, some are not. We just prove that there exists a series like above.

miracrown · Answer

Yeah, I'm afraid I won't be of much help. I'd need an hour or two to brush up on my algebra to be able to reasonably attempt this. Sowwyy :'/

loser66 · Answer

It's ok, friend. This is my homework and I am allowed to revise it if I don't get something at the first try. :) 
Thanks for your time.

miracrown · Answer

Yw :^)

dan815 · Answer

noob

loser66 · Answer

@Zarkon  give me your opinion, please