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Plus, to the proof any p-group G is solvable,although Z(G) is normal to G and which is abelian, but G/Z(G) is not abelian, so what is the chain subgroup to show that G is solvable?

6666
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Hint:

$$H\;\;\text{solvable}\;\implies\;\exists\;\;\text{an abelian series}\;\;1:=H_0\lhd H_1\lhd\ldots\lhd H_k:=H$$

$$G/H\;\;\text{solvable}\;\implies\;\exists\;\;\text{an abelian series}\;\;\overline 1:=\overline G_0\lhd \overline G_1\lhd\ldots\lhd\overline G_m:=G/H$$

Use that $\;G_i=N_i/H\;$ , with $\;H\le N_i\le G\;$ , and $\;\overline G_i\lhd\overline G_{i+1}\iff N_i\lhd N_{i+1}$

Now check what happens with the series

$$1=H_0\lhd\ldots\lhd H\lhd N_1\lhd\ldots ...\lhd G$$

Timbuc
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    (N.B.: OP will have to use $\bar G/\bar H\simeq G/H$) – Pedro Sep 16 '14 at 02:57
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    Why they are abelian series? By the definition, only the quotients are abelian. – 6666 Sep 16 '14 at 03:11
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    @Joseph, that's exactly what abelian series means: the factors are abelian. – Timbuc Sep 16 '14 at 03:12
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    But H is solvable, doesn't mean it's abelian. – 6666 Sep 16 '14 at 03:18
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    I know that, @Joseph...so what? – Timbuc Sep 16 '14 at 03:19
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    So do you mean H is also abelian? Or why you say by factor is abelian than the chain is abelian? – 6666 Sep 16 '14 at 03:20
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    No @Joseph, I don't say $;H;$ is abelian. And again: a (subnormal) series, or chain if you prefer, is abelian if each factor in it is abelian, meaning : $;H_i/H_{i-1};,;;\overline G_i/\overline G_{i-1};$ are abelian. – Timbuc Sep 16 '14 at 03:22
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    But how to prove the existence of these Ni? – 6666 Sep 16 '14 at 03:40
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    The correspondence theorem, @Joseph... this must be old stuff in group theory for you if you already are studying solvable groups. – Timbuc Sep 16 '14 at 03:45
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    Thank you very much, I am taking the modern algebra course, but I haven't seen the corresponding theorem, sorry about that. But right now I got it on line. – 6666 Sep 16 '14 at 03:53
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    That's odd, @Joseph, as the Correspondence Theorem is one of the most important, basic theorems in elementary group theory and, as far as I can see it, it preceedes solvable groups big time. – Timbuc Sep 16 '14 at 04:00
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    So did you learn the solvable group from modern algebra or other algebra courses? – 6666 Sep 16 '14 at 04:12
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    In a basic group theory course, which also included a little basic ring theory, @Joseph . – Timbuc Sep 16 '14 at 04:14