Commensurator of subgroup is subgroup: Difference between revisions

Latest revision as of 23:28, 29 April 2022

Statement

Suppose $H$ is a subgroup of a group $G$ . Consider the commensurator $K$ of $H$ in $G$ , defined as the set of all $g \in G$ such that $H \cap g H g^{- 1}$ is a subgroup of finite index in both $H$ and $g H g^{- 1}$ , i.e., $H$ and $g H g^{- 1}$ are commensurable subgroups. Then, $K$ is a subgroup of $G$ .

Facts used

Proof

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Given: A group $G$ , a subgroup $H$ of $G$ . $K$ is the set of all $g \in G$ such that $H \cap g H g^{- 1}$ has finite index in both $H$ and $g H g^{- 1}$ .

To prove: $K$ is a subgroup of $G$ .

Proof:

Proof for identity element

Let $e$ denote the identity element of $G$ . We have $e H e^{- 1} = H$ , so the intersection $H \cap e H e^{- 1}$ also equals $H$ . This has index $1$ in both $H$ and $e H e^{- 1}$ , which is finite.

Proof for inverses

Additional given: $g \in K$ . In other words, $H \cap g H g^{- 1}$ has finite index in both $H$ and $g H g^{- 1}$ .

To prove: $g^{- 1} \in K$ , i.e., $H \cap g^{- 1} H g$ has finite index in both $H$ and $g^{- 1} H g$ .

Proof

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	Consider the map $σ : G \to G$ given by $σ (x) : = g^{- 1} x g$ . This is an inner automorphism of $G$	Fact (1)	--	--
2	$σ$ preserves intersections and index of subgroups	Follows from definition of automorphism		Step (1)
3	$σ (H) = g^{- 1} H g$ and $σ (g H g^{- 1}) = H$ .			Step (1)
4	$σ (H \cap g H g^{- 1}) = g^{- 1} H g \cap H$ .			Steps (2), (3)
5	$H \cap g H g^{- 1}$ has finite index in both $H$ and $g H g^{- 1}$ .		$g \in K$ , the commensurator of $H$ .
6	$σ (H \cap g H g^{- 1}) = g^{- 1} H g \cap H$ has finite index in both $σ (H) = g^{- 1} H g$ and $σ (g H g^{- 1}) = H$ .			Steps (2), (3), (4), (5)	[SHOW MORE] We apply $σ$ to the statement of Step (5), which is permissible because automorphisms preserve index of subgroups. We use Steps (3) and (4) to compute what the images of the three subgroups under $σ$ .
7	$g^{- 1}$ is in $K$ .		definition of $K$	Step (6)	Step-definition direct.

Proof for products

Additional given: $g_{1}, g_{2} \in K$

To prove: $g_{1} g_{2} \in K$ , i.e., $H \cap (g_{1} g_{2}) H (g_{1} g_{2})^{- 1}$ has finite index in both $H$ and in $(g_{1} g_{2}) H (g_{1} g_{2})^{- 1}$ .

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	Consider the map $τ : G \to G$ given by $τ (x) = g_{1} x g_{1}^{- 1}$ . Then, $τ$ is an inner automorphism and in particular an automorphism of $G$ .	Fact (1)
2	$τ$ preserves intersections and index of subgroups	follows from definition of automorphism		Step (1)
3	$τ (H) = g_{1} H g_{1}^{- 1}$ and $τ (g_{2} H g_{2}^{- 1}) = g_{1} g_{2} H g_{2}^{- 1} g_{1}^{- 1} = (g_{1} g_{2}) H (g_{1} g_{2})^{- 1}$ .			Step (1)
4	$τ (H \cap g_{2} H g_{2}^{- 1}) = g_{1} H g_{1}^{- 1} \cap (g_{1} g_{2}) H (g_{1} g_{2})^{- 1}$ .			Steps (2), (3)
5	$H \cap g_{2} H g_{2}^{- 1}$ has finite index in both $H$ and $g_{2} H g_{2}^{- 1}$ .		$g_{2} \in K$ , definition of $K$		Given-direct
6	$g_{1} H g_{1}^{- 1} \cap (g_{1} g_{2}) H (g_{1} g_{2})^{- 1}$ has finite index in (i) $g_{1} H g_{1}^{- 1}$ and (ii) $(g_{1} g_{2}) H (g_{1} g_{2})^{- 1}$ .			Steps (2), (3), (4), (5)	[SHOW MORE] We apply $τ$ to both sides of Step (5), permissible by Step (2), and then simplify using Steps (3) and (4).
7	$H \cap (g_{1} H g_{1}^{- 1} \cap (g_{1} g_{2}) H (g_{1} g_{2})^{- 1})$ has finite index in (i) $H \cap g_{1} H g_{1}^{- 1}$ and (ii) $H \cap (g_{1} g_{2}) H (g_{1} g_{2})^{- 1}$ .	Fact (2)		Step (6)	[SHOW MORE] Intersect all terms of Step (6) with $H$ and apply Fact (2).
8	$H \cap g_{1} H g_{1}^{- 1}$ has finite index in (i) $H$ and in (ii) $g_{1} H g_{1}^{- 1}$		$g_{1} \in K$ , definition of $K$		Given-direct
9	$H \cap (g_{1} H g_{1}^{- 1} \cap (g_{1} g_{2}) H (g_{1} g_{2})^{- 1})$ has finite index in $H$ .	Fact (3)		Steps (7)(i), (8)(i)	[SHOW MORE] Combine Steps (7)(i) and (8)(i) with Fact (3) for the chain $H \cap (g_{1} H g_{1}^{- 1} \cap (g_{1} g_{2}) H (g_{1} g_{2})^{- 1}) \leq H \cap g_{1} H g_{1}^{- 1} \leq H$
10	$H \cap (g_{1} g_{2}) H (g_{1} g_{2})^{- 1}$ has finite index in $H$	Fact (3)		Step (9)	[SHOW MORE] Follows from Step (9), when we note that $H \cap (g_{1} g_{2}) H (g_{1} g_{2})^{- 1}$ is an intermediate subgroup between $H \cap (g_{1} H g_{1}^{- 1} \cap (g_{1} g_{2}) H (g_{1} g_{2})^{- 1})$ and $H$ .
11	$(H \cap g_{1} H g_{1}^{- 1}) \cap (g_{1} g_{2}) H (g_{1} g_{2})^{- 1}$ has finite index in (i) $H \cap (g_{1} g_{2}) H (g_{1} g_{2})^{- 1}$ and (ii) $g_{1} H g_{1}^{- 1} \cap (g_{1} g_{2}) H (g_{1} g_{2})^{- 1}$	Fact (2)		Step (8)	[SHOW MORE] Intersect all terms of Step (8) with $(g_{1} g_{2}) H (g_{1} g_{2})^{- 1}$ .
12	$(H \cap g_{1} H g_{1}^{- 1}) \cap (g_{1} g_{2}) H (g_{1} g_{2})^{- 1}$ has finite index in $(g_{1} g_{2}) H (g_{1} g_{2})^{- 1}$	Fact (3)		Steps (6)(ii), (11)(ii)	[SHOW MORE] Combine Steps (11)(ii) and (6)(ii) with Fact (4) for the chain $(H \cap g_{1} H g_{1}^{- 1}) \cap (g_{1} g_{2}) H (g_{1} g_{2})^{- 1} \leq g_{1} H g_{1}^{- 1} \cap (g_{1} g_{2}) H (g_{1} g_{2})^{- 1} \leq (g_{1} g_{2}) H (g_{1} g_{2})^{- 1}$ .
13	$H \cap (g_{1} g_{2}) H (g_{1} g_{2})^{- 1}$ has finite index in $(g_{1} g_{2}) H (g_{1} g_{2})^{- 1}$	Fact (3)		Step (12)	[SHOW MORE] Follows from Step (12), when we note that $H \cap (g_{1} g_{2}) H (g_{1} g_{2})^{- 1}$ is an intermediate subgroup between $(H \cap g_{1} H g_{1}^{- 1}) \cap (g_{1} g_{2}) H (g_{1} g_{2})^{- 1}$ and $(g_{1} g_{2}) H (g_{1} g_{2})^{- 1}$ .
14	$g_{1} g_{2} \in K$ , i.e., $H \cap (g_{1} g_{2}) H (g_{1} g_{2})^{- 1}$ has finite index in both $H$ and $(g_{1} g_{2}) H (g_{1} g_{2})^{- 1}$ .			Steps (10), (13)	Step-combination direct.

@@ Line 1: / Line 1: @@
+[[Page class::Fact| ]][[Difficulty level::3| ]]
 ==Statement==
-Suppose <math>H</math> is a [[subgroup]] of a [[group]] <math>G</math>. Consider the [[fact about::commensurator of a subgroup|commensurator]] <math>K</math> of <math>H</math> in <math>G</math>, defined as the set of all <math>g \in G</math> such that <math>H \cap gHg^{-1}</math> is a [[subgroup of finite index]] in both <math>H</math> and <math>gHg^{-1}</math>, i.e., <math>H</math> and <math>gHg^{-1}</math> are [[fact about::commensurable subgroups]]. Then, <math>K</math> is a subgroup of <math>G</math>.
+Suppose <math>H</math> is a [[subgroup]] of a [[group]] <math>G</math>. Consider the [[fact about::commensurator of a subgroup|commensurator]] <math>K</math> of <math>H</math> in <math>G</math>, defined as the set of all <math>g \in G</math> such that <math>H \cap gHg^{-1}</math> is a [[subgroup of finite index]] in both <math>H</math> and <math>gHg^{-1}</math>, i.e., <math>H</math> and <math>gHg^{-1}</math> are [[fact about::commensurable subgroups;2| ]][[commensurable subgroups]]. Then, <math>K</math> is a subgroup of <math>G</math>.
+==Facts used==
+# [[uses::Group acts as automorphisms by conjugation]]
+# [[uses::Index satisfies transfer inequality]]
+# [[uses::Index is multiplicative]]
 ==Proof==
+{{tabular proof format}}
 '''Given''': A group <math>G</math>, a subgroup <math>H</math> of <math>G</math>. <math>K</math> is the set of all <math>g \in G</math> such that <math>H \cap gHg^{-1}</math> has finite index in both <math>H</math> and <math>gHg^{-1}</math>.
 '''To prove''': <math>K</math> is a subgroup of <math>G</math>.
+'''Proof''':
+===Proof for identity element===
+Let <math>e</math> denote the identity element of <math>G</math>. We have <math>eHe^{-1} = H</math>, so the intersection <math>H \cap eHe^{-1}</math> also equals <math>H</math>. This has index <math>1</math> in both <math>H</math> and <math>eHe^{-1}</math>, which is finite.
+===Proof for inverses===
+'''Additional given''': <math>g \in K</math>. In other words, <math>H \cap gHg^{-1}</math> has finite index in both <math>H</math> and <math>gHg^{-1}</math>.
+'''To prove''': <math>g^{-1} \in K</math>, i.e., <math>H \cap g^{-1}Hg</math> has finite index in both <math>H</math> and <math>g^{-1}Hg</math>.
+'''Proof'''
+{| class="sortable" border="1"
+! Step no. !! Assertion/construction !! Facts used !! Given data used !! Previous steps used !! Explanation
+|-
+| 1 || Consider the map <math>\sigma:G \to G</math> given by <math>\sigma(x) := g^{-1}xg</math>. This is an inner automorphism of <math>G</math> || Fact (1) || -- || -- ||
+|-
+| 2 || <math>\sigma</math> preserves intersections and index of subgroups || Follows from definition of automorphism || || Step (1) ||
+|-
+| 3 || <math>\sigma(H) = g^{-1}Hg</math> and <math>\sigma(gHg^{-1}) = H</math>. || || || Step (1) ||
+|-
+| 4 || <math>\sigma(H \cap gHg^{-1}) = g^{-1}Hg \cap H</math>. || || || Steps (2), (3) ||
+|-
+| 5 || <math>H \cap gHg^{-1}</math> has finite index in both <math>H</math> and <math>gHg^{-1}</math>. || || <math>g \in K</math>, the commensurator of <math>H</math>. || ||
+|-
+| 6 || <math>\sigma(H \cap gHg^{-1}) = g^{-1}Hg \cap H</math> has finite index in both <math>\! \sigma(H) = g^{-1}Hg</math> and <math>\! \sigma(gHg^{-1}) = H</math>. || || || Steps (2), (3), (4), (5) || <toggledisplay>We apply <math>\sigma</math> to the statement of Step (5), which is permissible because automorphisms preserve index of subgroups. We use Steps (3) and (4) to compute what the images of the three subgroups under <math>\sigma</math>.</toggledisplay>
+|-
+| 7 || <math>g^{-1}</math> is in <math>K</math>. || || definition of <math>K</math> || Step (6) || Step-definition direct.
+|}
+===Proof for products===
+'''Additional given''': <math>g_1, g_2 \in K</math>
+'''To prove''': <math>g_1g_2 \in K</math>, i.e., <math>H \cap (g_1g_2)H(g_1g_2)^{-1}</math> has finite index in both <math>H</math> and in <math>(g_1g_2)H(g_1g_2)^{-1}</math>.
 '''Proof''':
 {| class="sortable" border="1"
-! Step no. !! Assertion/construction !! Explanation
+! Step no. !! Assertion/construction !! Facts used !! Given data used !! Previous steps used !! Explanation
+|-
+| 1 || Consider the map <math>\tau:G \to G</math> given by <math>\tau(x) = g_1xg_1^{-1}</math>. Then, <matH>\tau</math> is an inner automorphism and in particular an automorphism of <math>G</math>. || Fact (1) || || ||
+|-
+| 2 || <math>\tau</math> preserves intersections and index of subgroups || follows from definition of automorphism || || Step (1) ||
+|-
+| 3 || <math>\tau(H) = g_1Hg_1^{-1}</math> and <math>\tau(g_2Hg_2^{-1}) = g_1g_2Hg_2^{-1}g_1^{-1} = (g_1g_2)H(g_1g_2)^{-1}</math>. || || || Step (1) ||
+|-
+| 4 || <math>\tau(H \cap g_2Hg_2^{-1}) = g_1Hg_1^{-1} \cap (g_1g_2)H(g_1g_2)^{-1}</math>. || || || Steps (2), (3) ||
+|-
+| 5 ||  <math>H \cap g_2Hg_2^{-1}</math> has finite index in both <math>\! H</math> and <math>\! g_2Hg_2^{-1}</math>. || || <math>g_2 \in K</math>, definition of <math>K</math> || || Given-direct
+|-
+| 6 || <math>g_1Hg_1^{-1} \cap (g_1g_2)H(g_1g_2)^{-1}</math> has finite index in (i) <math>\! g_1Hg_1^{-1}</math> and (ii) <math>\! (g_1g_2)H(g_1g_2)^{-1}</math>. || || || Steps (2), (3), (4), (5) || <toggledisplay>We apply <math>\tau</math> to both sides of Step (5), permissible by Step (2), and then simplify using Steps (3) and (4).</toggledisplay>
+|-
+| 7 || <math>H \cap (g_1Hg_1^{-1} \cap (g_1g_2)H(g_1g_2)^{-1})</math> has finite index in (i) <math>H \cap g_1Hg_1^{-1}</math> and (ii) <math>H \cap (g_1g_2)H(g_1g_2)^{-1}</math>. || Fact (2) || || Step (6) || <toggledisplay>Intersect all terms of Step (6) with <math>H</math> and apply Fact (2).</toggledisplay>
+|-
+| 8 || <math>H \cap g_1Hg_1^{-1}</math> has finite index in (i) <math>\!H</math> and in (ii) <math>g_1Hg_1^{-1}</math> || || <math>g_1 \in K</math>, definition of <math>K</math> || || Given-direct
+|-
+| 9 || <math>H \cap (g_1Hg_1^{-1} \cap (g_1g_2)H(g_1g_2)^{-1})</math> has finite index in <math>\! H</math>. || Fact (3) || || Steps (7)(i), (8)(i) || <toggledisplay>Combine Steps (7)(i) and (8)(i) with Fact (3) for the chain <math>H \cap (g_1Hg_1^{-1} \cap (g_1g_2)H(g_1g_2)^{-1}) \le H \cap g_1Hg_1^{-1} \le H</math></toggledisplay>
+|-
+| 10 || <math>H \cap (g_1g_2)H(g_1g_2)^{-1}</math> has finite index in <math>\! H</math> || Fact (3) || || Step (9) || <toggledisplay>Follows from Step (9), when we note that <math>H \cap (g_1g_2)H(g_1g_2)^{-1}</math> is an intermediate subgroup between <math>H \cap (g_1Hg_1^{-1} \cap (g_1g_2)H(g_1g_2)^{-1})</math> and <math>H</math>.</toggledisplay>
+|-
+| 11 || <math>(H \cap g_1Hg_1^{-1}) \cap (g_1g_2)H(g_1g_2)^{-1}</math> has finite index in (i) <math>H \cap (g_1g_2)H(g_1g_2)^{-1}</math> and (ii) <math>g_1Hg_1^{-1} \cap (g_1g_2)H(g_1g_2)^{-1}</math> || Fact (2) || || Step (8) || <toggledisplay>Intersect all terms of Step (8) with <math>(g_1g_2)H(g_1g_2)^{-1}</math>.</toggledisplay>
 |-
-| 1 || The identity element <math>e</math> of <math>G</math> is in <math>K</math> || <toggledisplay><math>eHe^{-1} = H</math>, so the intersection <math>H \cap eHe^{-1}</math> also equals <math>H</math>. This has index <math>1</math> in both <math>H</math> and <math>eHe^{-1}</math>, which is finite.</toggledisplay>
+| 12 || <math>(H \cap g_1Hg_1^{-1}) \cap (g_1g_2)H(g_1g_2)^{-1}</math> has finite index in <math>\! (g_1g_2)H(g_1g_2)^{-1}</math> || Fact (3) || || Steps (6)(ii), (11)(ii) || <toggledisplay>Combine Steps (11)(ii) and (6)(ii) with Fact (4) for the chain <math>(H \cap g_1Hg_1^{-1}) \cap (g_1g_2)H(g_1g_2)^{-1} \le g_1Hg_1^{-1} \cap (g_1g_2)H(g_1g_2)^{-1} \le (g_1g_2)H(g_1g_2)^{-1}</math>.</toggledisplay>
 |-
-| 2 || If <math>g \in K</math>, then <math>g^{-1} \in K</math> || <toggledisplay>Consider the map <math>\sigma:G \to G</math> given by <math>x \mapsto g^{-1}xg</math>. This is an inner automorphism of <math>G</math>, and hence preserves intersections and the index of subgroups. We have <math>\sigma(H) = g^{-1}Hg</math> and <math>\sigma(gHg^{-1}) = H</math>. Thus, we get <math>\sigma(H \cap gHg^{-1}) = g^{-1}Hg \cap H</math>. Since the index of <math>H \cap gHg^{-1}</math> in both <math>H</math> and <math>gHg^{-1}</math> is finite, so is the index of <math>\sigma(H \cap gHg^{-1}) = g^{-1}Hg \cap H</math> in both <math>\sigma(H) = g^{-1}Hg</math> and <math>\sigma(gHg^{-1}) = H</math>.</toggledisplay>
+| 13 || <math>H \cap (g_1g_2)H(g_1g_2)^{-1}</math> has finite index in <math>\! (g_1g_2)H(g_1g_2)^{-1}</math> || Fact (3) || || Step (12) || <toggledisplay>Follows from Step (12), when we note that <math>H \cap (g_1g_2)H(g_1g_2)^{-1}</math> is an intermediate subgroup between <math>(H \cap g_1Hg_1^{-1}) \cap (g_1g_2)H(g_1g_2)^{-1}</math> and <math>(g_1g_2)H(g_1g_2)^{-1}</math>.</toggledisplay>
 |-
-| 3 || If <math>g_1,g_2 \in K</math>, then <math>g_1g_2 \in K</math>. || {{fillin}}
+| 14 || <math>g_1g_2 \in K</math>, i.e., <math>H \cap (g_1g_2)H(g_1g_2)^{-1}</math> has finite index in both <math>H</math> and <math>\! (g_1g_2)H(g_1g_2)^{-1}</math>.  || || || Steps (10), (13) || Step-combination direct.
 |}