# Infinite length Composition series

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 $i >N$.

• Clearly the group has to be infinite for the problem not to be boringly trivial, so your first parentheses is odd. – DonAntonio Jul 12 '16 at 21:30
• @DonAntonio I just wanted people not to get a wrong idea. – Primo Jul 12 '16 at 21:32
• This is the same question as math.stackexchange.com/questions/1844155 – Derek Holt Jul 12 '16 at 22:04
• @DerekHolt I don't think the question is the same. I am assuming $G$ has at least one composition series. – Primo Jul 12 '16 at 22:07
• It is the identical questions worded differently. – Derek Holt Jul 13 '16 at 6:33

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.
• I am assuming $G$ has a composition series. Does your example have a composition series? – Primo Jul 12 '16 at 21:34
• Yes, $G_{i+1}$ is normal in $G_i$ – Tsemo Aristide Jul 12 '16 at 21:37