Commensurator of subgroup is subgroup

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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 gHg^{-1} is a subgroup of finite index in both H and gHg^{-1}, i.e., H and gHg^{-1} are Commensurable subgroups (?). Then, K is a subgroup of G.

Facts used

  1. Group acts as automorphisms by conjugation
  2. Index satisfies transfer inequality
  3. Index is multiplicative

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 gHg^{-1} has finite index in both H and gHg^{-1}.

To prove: K is a subgroup of G.

Proof:

Proof for identity element

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

Proof for inverses

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

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

Proof

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 Consider the map \sigma:G \to G given by \sigma(x) := g^{-1}xg. This is an inner automorphism of G Fact (1) -- --
2 \sigma preserves intersections and index of subgroups Follows from definition of automorphism Step (1)
3 \sigma(H) = g^{-1}Hg and \sigma(gHg^{-1}) = H. Step (1)
4 \sigma(H \cap gHg^{-1}) = g^{-1}Hg \cap H. Steps (2), (3)
5 H \cap gHg^{-1} has finite index in both H and gHg^{-1}. g \in K, the commensurator of H.
6 \sigma(H \cap gHg^{-1}) = g^{-1}Hg \cap H has finite index in both \! \sigma(H) = g^{-1}Hg and \! \sigma(gHg^{-1}) = H. Steps (2), (3), (4), (5) [SHOW MORE]
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_1g_2 \in K, i.e., H \cap (g_1g_2)H(g_1g_2)^{-1} has finite index in both H and in (g_1g_2)H(g_1g_2)^{-1}.

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 Consider the map \tau:G \to G given by \tau(x) = g_1xg_1^{-1}. Then, \tau is an inner automorphism and in particular an automorphism of G. Fact (1)
2 \tau preserves intersections and index of subgroups follows from definition of automorphism Step (1)
3 \tau(H) = g_1Hg_1^{-1} and \tau(g_2Hg_2^{-1}) = g_1g_2Hg_2^{-1}g_1^{-1} = (g_1g_2)H(g_1g_2)^{-1}. Step (1)
4 \tau(H \cap g_2Hg_2^{-1}) = g_1Hg_1^{-1} \cap (g_1g_2)H(g_1g_2)^{-1}. Steps (2), (3)
5 H \cap g_2Hg_2^{-1} has finite index in both \! H and \! g_2Hg_2^{-1}. g_2 \in K, definition of K Given-direct
6 g_1Hg_1^{-1} \cap (g_1g_2)H(g_1g_2)^{-1} has finite index in (i) \! g_1Hg_1^{-1} and (ii) \! (g_1g_2)H(g_1g_2)^{-1} Steps (2), (3), (4), (5) [SHOW MORE]
7 H \cap (g_1Hg_1^{-1} \cap (g_1g_2)H(g_1g_2)^{-1}) has finite index in (i) H \cap g_1Hg_1^{-1} and (ii) H \cap (g_1g_2)H(g_1g_2)^{-1}. Fact (2) Step (6) [SHOW MORE]
8 H \cap g_1Hg_1^{-1} has finite index in (i) \!H and in (ii) g_1Hg_1^{-1} g_1 \in K, definition of K Given-direct
9 H \cap (g_1Hg_1^{-1} \cap (g_1g_2)H(g_1g_2)^{-1}) has finite index in \! H. Fact (3) Steps (7)(i), (8)(i) [SHOW MORE]
10 H \cap (g_1g_2)H(g_1g_2)^{-1} has finite index in \! H Fact (3) Step (9) [SHOW MORE]
11 (H \cap g_1Hg_1^{-1}) \cap (g_1g_2)H(g_1g_2)^{-1} has finite index in (i) H \cap (g_1g_2)H(g_1g_2)^{-1} and (ii) g_1Hg_1^{-1} \cap (g_1g_2)H(g_1g_2)^{-1} Fact (2) Step (8) [SHOW MORE]
12 (H \cap g_1Hg_1^{-1}) \cap (g_1g_2)H(g_1g_2)^{-1} has finite index in \! (g_1g_2)H(g_1g_2)^{-1} Fact (3) Steps (6)(ii), (11)(ii) [SHOW MORE]
13 H \cap (g_1g_2)H(g_1g_2)^{-1} has finite index in \! (g_1g_2)H(g_1g_2)^{-1} Fact (3) Step (12) [SHOW MORE]
14 g_1g_2 \in K, i.e., H \cap (g_1g_2)H(g_1g_2)^{-1} has finite index in both H and \! (g_1g_2)H(g_1g_2)^{-1}. Steps (10), (13) Step-combination direct.