Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.