Every nontrivial subgroup of a solvable group contains normal abelian subgroups

Prove that if H is a nontrivial subgroup of a solvable group G then there is a nontrivial subgroup A \leq H which is abelian and normal in G.


We proceed by induction on the width of G– that is, the number of prime factors of |G| (including multiplicity).

If G has width 1, then G is simple and no such H exists.

If G has width 2, say |G| = pq, and H \leq G is nontrivial, then without loss of generality |H| = p.

Lemma: Let (\mathcal{P},\leq) be a nonempty partially ordered set such that every chain has bounded length. Then there exists a minimal element m \in \mathcal{P}. Proof: Let \{C_i\}_{i=1}^n \subseteq \mathcal{P} be a chain of maximal length n, where C_i \leq C_{i+1}. If an element C_0 \in \mathcal{P} exists such that C_0 \leq C_1, then \{ C_i \}_{i=0}^n is a chain of length n+1, a contradiction. So C_1 is \leq-minimal in \mathcal{P}. \square

Consider the set \mathcal{H} = \{ A \vartriangleleft G \ |\ A \leq H, A \neq 1 \}. Note that \mathcal{H} is nonempty since H \in \mathcal{H}. Note that because G is solvable, G has a finite composition series, so that by the Jordan-Holder Theorem, all composition series for G have the same finite length n. Now every chain in \mathcal{H} can be extended to a composition series for G, and thus has length bounded by n. By the lemma, then, there exists a \leq-minimal element A \in \mathcal{H}. By Lemma 5 to this previous theorem, A is abelian.

Advertisements
Post a comment or leave a trackback: Trackback URL.

Comments

  • steve  On August 3, 2011 at 8:41 pm

    “Note that because G is solvable, G has a finite composition series.” Can this be right? The problem doesn’t assume G is finite, and Z is a solvable group that doesn’t have a composition series.

    • nbloomf  On August 5, 2011 at 10:18 am

      Ah, this proof implicitly assumes that G is finite. I’ll have to mark this for later.

      Thanks!

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: