Namespaces
Variants
Actions

Difference between revisions of "Normalizer of a subset"

From Encyclopedia of Mathematics
Jump to: navigation, search
(Importing text file)
 
m (tex encoded by computer)
 
Line 1: Line 1:
''<img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/n/n067/n067740/n0677401.png" /> of a [[Group|group]] <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/n/n067/n067740/n0677402.png" /> in a subgroup <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/n/n067/n067740/n0677403.png" /> of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/n/n067/n067740/n0677404.png" />''
+
<!--
 +
n0677401.png
 +
$#A+1 = 29 n = 0
 +
$#C+1 = 29 : ~/encyclopedia/old_files/data/N067/N.0607740 Normalizer of a subset
 +
Automatically converted into TeX, above some diagnostics.
 +
Please remove this comment and the {{TEX|auto}} line below,
 +
if TeX found to be correct.
 +
-->
 +
 
 +
{{TEX|auto}}
 +
{{TEX|done}}
 +
 
 +
'' $  M $
 +
of a [[Group|group]] $  G $
 +
in a subgroup $  H $
 +
of $  G $''
  
 
The set
 
The set
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/n/n067/n067740/n0677405.png" /></td> </tr></table>
+
$$
 +
N _ {H} ( M)  = \
 +
\{ {h } : {h \in H , h  ^ {-} 1 M h = M } \}
 +
,
 +
$$
  
that is, the set of all elements <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/n/n067/n067740/n0677406.png" /> of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/n/n067/n067740/n0677407.png" /> such that <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/n/n067/n067740/n0677408.png" /> (the conjugate of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/n/n067/n067740/n0677409.png" /> by <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/n/n067/n067740/n06774010.png" />) for every <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/n/n067/n067740/n06774011.png" /> also belongs to <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/n/n067/n067740/n06774012.png" />. For any <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/n/n067/n067740/n06774013.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/n/n067/n067740/n06774014.png" /> the normalizer <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/n/n067/n067740/n06774015.png" /> is a subgroup of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/n/n067/n067740/n06774016.png" />. An important special case is the normalizer of a subgroup of a group <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/n/n067/n067740/n06774017.png" /> in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/n/n067/n067740/n06774018.png" />. A subgroup <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/n/n067/n067740/n06774019.png" /> of a group <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/n/n067/n067740/n06774020.png" /> is normal (or invariant, cf. [[Invariant subgroup|Invariant subgroup]]) in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/n/n067/n067740/n06774021.png" /> if and only if <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/n/n067/n067740/n06774022.png" />. The normalizer of a set consisting of a single element is the same as its [[Centralizer|centralizer]]. For any <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/n/n067/n067740/n06774023.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/n/n067/n067740/n06774024.png" /> the cardinality of the class of subsets conjugate to <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/n/n067/n067740/n06774025.png" /> by elements of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/n/n067/n067740/n06774026.png" /> (that is, subsets of the form <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/n/n067/n067740/n06774027.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/n/n067/n067740/n06774028.png" />) is equal to the index <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/n/n067/n067740/n06774029.png" />.
+
that is, the set of all elements $  h $
 +
of $  H $
 +
such that $  h  ^ {-} 1 m h $(
 +
the conjugate of $  m $
 +
by $  h $)  
 +
for every $  m \in M $
 +
also belongs to $  M $.  
 +
For any $  M $
 +
and $  H $
 +
the normalizer $  N _ {H} ( M) $
 +
is a subgroup of $  H $.  
 +
An important special case is the normalizer of a subgroup of a group $  G $
 +
in $  G $.  
 +
A subgroup $  A $
 +
of a group $  G $
 +
is normal (or invariant, cf. [[Invariant subgroup|Invariant subgroup]]) in $  G $
 +
if and only if $  N _ {G} ( A) = G $.  
 +
The normalizer of a set consisting of a single element is the same as its [[Centralizer|centralizer]]. For any $  H $
 +
and $  M $
 +
the cardinality of the class of subsets conjugate to $  M $
 +
by elements of $  H $(
 +
that is, subsets of the form $  h  ^ {-} 1 M h $,  
 +
$  h \in H $)  
 +
is equal to the index $  | H : N _ {H} ( M) | $.
  
 
====References====
 
====References====
 
<table><TR><TD valign="top">[1]</TD> <TD valign="top">  M.I. Kargapolov,  J.I. [Yu.I. Merzlyakov] Merzljakov,  "Fundamentals of the theory of groups" , Springer  (1979)  (Translated from Russian)</TD></TR></table>
 
<table><TR><TD valign="top">[1]</TD> <TD valign="top">  M.I. Kargapolov,  J.I. [Yu.I. Merzlyakov] Merzljakov,  "Fundamentals of the theory of groups" , Springer  (1979)  (Translated from Russian)</TD></TR></table>
 
 
  
 
====Comments====
 
====Comments====
 
  
 
====References====
 
====References====
 
<table><TR><TD valign="top">[a1]</TD> <TD valign="top">  D.J.S. Robinson,  "A course in the theory of groups" , Springer  (1980)</TD></TR></table>
 
<table><TR><TD valign="top">[a1]</TD> <TD valign="top">  D.J.S. Robinson,  "A course in the theory of groups" , Springer  (1980)</TD></TR></table>

Latest revision as of 08:03, 6 June 2020


$ M $ of a group $ G $ in a subgroup $ H $ of $ G $

The set

$$ N _ {H} ( M) = \ \{ {h } : {h \in H , h ^ {-} 1 M h = M } \} , $$

that is, the set of all elements $ h $ of $ H $ such that $ h ^ {-} 1 m h $( the conjugate of $ m $ by $ h $) for every $ m \in M $ also belongs to $ M $. For any $ M $ and $ H $ the normalizer $ N _ {H} ( M) $ is a subgroup of $ H $. An important special case is the normalizer of a subgroup of a group $ G $ in $ G $. A subgroup $ A $ of a group $ G $ is normal (or invariant, cf. Invariant subgroup) in $ G $ if and only if $ N _ {G} ( A) = G $. The normalizer of a set consisting of a single element is the same as its centralizer. For any $ H $ and $ M $ the cardinality of the class of subsets conjugate to $ M $ by elements of $ H $( that is, subsets of the form $ h ^ {-} 1 M h $, $ h \in H $) is equal to the index $ | H : N _ {H} ( M) | $.

References

[1] M.I. Kargapolov, J.I. [Yu.I. Merzlyakov] Merzljakov, "Fundamentals of the theory of groups" , Springer (1979) (Translated from Russian)

Comments

References

[a1] D.J.S. Robinson, "A course in the theory of groups" , Springer (1980)
How to Cite This Entry:
Normalizer of a subset. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Normalizer_of_a_subset&oldid=48020
This article was adapted from an original article by N.N. Vil'yams (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article