View source for There exist subgroups of arbitrarily large subnormal depth

{{arbitrarily large subnormal depth statement}}

==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 subgroup]] of <math>G</math> but is ''not'' a <math>(k-1)</math>-subnormal subgroup of <math>G</math>. In other words, the subnormal depth of <math>H</math> in <math>G</math> is precisely <math>k</math>. Equivalently, there exists a series of subgroups:

<math>H = H_0 \le H_1 \le H_2 \le \dots \le H_k = G</math>

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.

==Related facts==

* [[Stronger than::Normality is not transitive]]
* [[Weaker than::Descendant not implies subnormal]], [[Weaker than::there exist subgroups of arbitrarily large descendant depth]]
* [[Weaker than::Ascendant not implies subnormal]], [[Weaker than::there exist subgroups of arbitrarily large ascendant depth]]
* [[Weaker than::Normal not implies left-transitively fixed-depth subnormal]]
* [[Weaker than::Normal not implies right-transitively fixed-depth subnormal]]

==Proof==

===Example of the dihedral group===

{{further|[[Particular example::dihedral group]]}}

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

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

Let <math>H</math> be the two-element subgroup generated by <math>x</matH>:

<math>H = \langle x \rangle = \{ e, x \}</math>.

* <math>H</math> is <math>k</math>-subnormal in <math>G</math>. Consider the series:

<math>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</math>.

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

* <math>H</math> is not <math>(k-1)</math>-subnormal in <math>G</math>: 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 <math>H</math> in the group to its right.