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Let $G$ be a group (possibly infinite). Suppose $G$ has a composition series. I could show that any other composition series has the same length. But I cannot prove the following.

Let $G \triangleright G_2 \triangleright G_3 \triangleright \cdots$ be a series of normal groups (may or may not assume the successive quotients are simple). I want to show that this series should terminate, i.e., there is $N$ such that $G_i=G_N$ for any $i >N$.

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    $\begingroup$ Clearly the group has to be infinite for the problem not to be boringly trivial, so your first parentheses is odd. $\endgroup$ – DonAntonio Jul 12 '16 at 21:30
  • $\begingroup$ @DonAntonio I just wanted people not to get a wrong idea. $\endgroup$ – Primo Jul 12 '16 at 21:32
  • $\begingroup$ This is the same question as math.stackexchange.com/questions/1844155 $\endgroup$ – Derek Holt Jul 12 '16 at 22:04
  • $\begingroup$ @DerekHolt I don't think the question is the same. I am assuming $G$ has at least one composition series. $\endgroup$ – Primo Jul 12 '16 at 22:07
  • $\begingroup$ It is the identical questions worded differently. $\endgroup$ – Derek Holt Jul 13 '16 at 6:33
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This is not true, take $G=Z^N$ and $G_i$ the subgroup such that $x\in G_i$ if its first $i$ components are zero.

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    $\begingroup$ I am assuming $G$ has a composition series. Does your example have a composition series? $\endgroup$ – Primo Jul 12 '16 at 21:34
  • $\begingroup$ Yes, $G_{i+1}$ is normal in $G_i$ $\endgroup$ – Tsemo Aristide Jul 12 '16 at 21:37
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    $\begingroup$ By a composition series, I mean a finite length series of normal groups whose successive quotients are simple. $\endgroup$ – Primo Jul 12 '16 at 21:38
  • $\begingroup$ If the serie is finite it terminates, ...you precised that you don't request to the quotients to be simple. $\endgroup$ – Tsemo Aristide Jul 12 '16 at 21:40
  • $\begingroup$ @TsemoAristide An abelian group has a (finite, by definition) composition series iff it is finite. $\endgroup$ – DonAntonio Jul 12 '16 at 21:40

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