Verbality is quotient-transitive: Difference between revisions

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Then, <math>K</matH> must be a verbal subgroup of <math>G</math>. In fact, the set of words that we can use to generate <math>K</math> is obtained by taking products of the set of words used for <math>H</math> in <math>G</math> and the set of words used for <math>K/H</math> in <math>G/H</math>.
Then, <math>K</matH> must be a verbal subgroup of <math>G</math>. In fact, the set of words that we can use to generate <math>K</math> is obtained by taking products of the set of words used for <math>H</math> in <math>G</math> and the set of words used for <math>K/H</math> in <math>G/H</math>.
==Related facts==
* [[Verbality is transitive]]

Latest revision as of 17:43, 17 July 2013

This article gives the statement, and possibly proof, of a subgroup property (i.e., verbal subgroup) satisfying a subgroup metaproperty (i.e., quotient-transitive subgroup property)
View all subgroup metaproperty satisfactions | View all subgroup metaproperty dissatisfactions |Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about verbal subgroup |Get facts that use property satisfaction of verbal subgroup | Get facts that use property satisfaction of verbal subgroup|Get more facts about quotient-transitive subgroup property


Statement

Suppose G is a group, with HKG subgroups. Suppose H is a verbal subgroup of G. Note that since verbal implies normal, H is a normal subgroup of G, so that we can talk of the quotient group G/H. Suppose we are further given that K/H is a verbal subgroup of G/H.

Then, K must be a verbal subgroup of G. In fact, the set of words that we can use to generate K is obtained by taking products of the set of words used for H in G and the set of words used for K/H in G/H.

Related facts