Variety-containing implies omega subgroup in group of prime power order

From Groupprops
Revision as of 22:33, 10 August 2009 by Vipul (talk | contribs) (Created page with '==Definition== Suppose <math>P</math> is a group of prime power order, i.e., a finite <math>p</math>-group for some prime number <math>p</math>. Suppose <math>H</math> i…')
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to: navigation, search

Definition

Suppose P is a group of prime power order, i.e., a finite p-group for some prime number p. Suppose H is a Subisomorph-containing subgroup (?) of P: a subgroup of P such that any subgroup of P isomorphic to a subgroup of H is itself contained in H. Then, H is one of the omega subgroups of P. In other words, there is some k such that:

H = \Omega_k(P) := \langle x \mid x^{p^k} = e \rangle.

Proof

Given: A finite p-group P, a subisomorph-containing subgroup H of P.

To prove: H = \Omega_k(P) for some natural number k.

Proof: Let p^k be the exponent of H. Then, we clearly have:

H \le \Omega_k(P) = \langle x \mid x^{p^k} = e \rangle.

Next, we show that H = \Omega_k(P). Suppose x \in P is such that x^{p^k} = e. Then, since p^k is the exponent of H, there exists y \in H such that the order of y is p^k. Suppose the order of x is p^l, l \le k. Then, \langle x \rangle \cong \langle y^{p^{k-l}} \rangle. Thus, \langle x \rangle is isomorphic to a subgroup of H, so \langle x \rangle \le H. Thus, \Omega_k(P) = H.