Difference between revisions of "There exist subgroups of arbitrarily large subnormal depth"

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(New page: ==Statement== Let <math>k</math> be a positive integer. Then, we can find a group <math>G</math> and a subgroup <math>H</math> such that <math>H</math> is a <math>k</math>-[[subnormal sub...)
 
(→‎Related facts)
 
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{{arbitrarily large subnormal depth statement}}
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==Statement==
 
==Statement==
  
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with each <math>H_i</math> normal in <math>H_{i+1}</math>, and there exists ''no'' series of length <math>k-1</math> with the property.
 
with each <math>H_i</math> normal in <math>H_{i+1}</math>, and there exists ''no'' series of length <math>k-1</math> with the property.
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==Related facts==
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* [[Stronger than::Normality is not transitive]]
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* [[Weaker than::Descendant not implies subnormal]], [[Weaker than::there exist subgroups of arbitrarily large descendant depth]]
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* [[Weaker than::Ascendant not implies subnormal]], [[Weaker than::there exist subgroups of arbitrarily large ascendant depth]]
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* [[Weaker than::Normal not implies left-transitively fixed-depth subnormal]]
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* [[Weaker than::Normal not implies right-transitively fixed-depth subnormal]]
  
 
==Proof==
 
==Proof==

Latest revision as of 22:41, 19 April 2009

This is a statement of the form: there exist subnormal subgroups of arbitrarily large subnormal depth satisfying certain conditions.
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Statement

Let k be a positive integer. Then, we can find a group G and a subgroup H such that H is a k-subnormal subgroup of G but is not a (k-1)-subnormal subgroup of G. In other words, the subnormal depth of H in G is precisely k. Equivalently, there exists a series of subgroups:

H = H_0 \le H_1 \le H_2 \le \dots \le H_k = G

with each H_i normal in H_{i+1}, and there exists no series of length k-1 with the property.

Related facts

Proof

Example of the dihedral group

Further information: dihedral group

Let G be the dihedral group of order 2^{k+1}. Specifically, we have:

G = \langle a,x \mid a^{2^k} = x^2 = e, xax^{-1} = a^{-1} \rangle.

Let H be the two-element subgroup generated by x:

H = \langle x \rangle = \{ e, x \}.

  • H is k-subnormal in G. Consider the series:

H = \langle x \rangle \le \langle a^{2^{k-1}}, x \rangle \le \langle a^{2^{k-2}}, x \rangle \le \dots \le \langle a^2, x \rangle \le \langle a,x \rangle = G.

Each subgroup has index two in its predecessor, and is thus normal. The series has length k, so H is k-subnormal in G.

  • H is not (k-1)-subnormal in G: To see this, note that the above subnormal series is a subnormal series of minimum length, because, starting from the right, each subgroup is the normal closure of H in the group to its right.
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