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Prove that the property max-n, is not inherited by normal subgroups. [Let $A$ be the additive groups of rational numbers of the for $m2^n$, $m,n \in \mathbb{Z}$, and let $T = \langle t \rangle$ be infinite cyclic. Let $t$ act on $A$ by the rule $ta = 2a$. Now consider the group $G$ which is the semidirect product of $T$ and $A$.]

Here, a group has the property max-n means any set of normal subgroups of this group has a maximal element.

The group $A$ certainly does not have this property as $\langle \frac{1}{2} \rangle \subsetneq \langle \frac{1}{2^2} \rangle \subsetneq \cdots \subsetneq \langle \frac{1}{2^n} \rangle \subsetneq \cdots$ is an infinite ascending sequence of normal subgroups. Then, I have to prove $G$ has the property max-n.

First, suppose that the normal subgroup $N$ is contained in $A$ (the semidirect product of $\{1 \}$ and $A$). If $N$ is not trivial, then $(1,k) \in N$ for some integer $k$. As $(t,0)(1,k)(t,0)^{-1} = (t,0)(1,k)(t^{-1},0) = (1,2^{-1}k)$, $\{ 2^nk | n \in \mathbb{Z} \} \subseteq N$. It is clear that any nontrivial normal subgroup of $G$ contained in $A$ never appears in an infinite ascending sequence. Then it remains to consider the normal subgroups which have nontrivial intersections with $T$. But the computation becomes more and mor complicate.

So, do I have to find out all the possible normal subgroups of $G$, in order to remove the possibility of infinite sequence? If this is necessary, is there any better idea of finding all these normal subgroups? If not, how can I prove the max-n property of $G$?

Many thanks.

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up vote 2 down vote accepted

The normal subgroups of $G$ contained in $A$ are all of the form $\{ k2^n : n \in \mathbb{Z}, k >0, k\ {\rm odd}\}$ so (as you observed in your post), an ascending chain of such subgroups must stabilize.

Now let $N_1 < N_2 < \cdots$ be an ascending chain of normal subgroups of $G$. Then $N_iA/A$ is an ascending chain of normal subgroups of an infinite cyclic group, so must stabilize. Also $N_i \cap A$ is an ascending chain of normal subgroups of $G$ contained in $A$, so it must stabilize. Once both of these chains have stabilized, so has $N_i$ itself.

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Thank you very much! If I am not mistaken, this is like the method of proving the Jordan–Hölder theorem. I forgot it. Thank you very much. – ShinyaSakai Oct 30 '11 at 15:44

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