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.

Proof

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:

Step no. Assertion/construction Explanation
1 The identity element e of G is in K 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.
2 If g \in K, then g^{-1} \in K Consider the map \sigma:G \to G given by x \mapsto g^{-1}xg. This is an inner automorphism of G, and hence preserves intersections and the index of subgroups. We have \sigma(H) = g^{-1}Hg and \sigma(gHg^{-1}) = H. Thus, we get \sigma(H \cap gHg^{-1}) = g^{-1}Hg \cap H. Since the index of H \cap gHg^{-1} in both H and gHg^{-1} is finite, so is the index of \sigma(H \cap gHg^{-1}) = g^{-1}Hg in both \sigma(H) = g^{-1}Hg and \sigma(gHg^{-1}) = H.
3 If g_1,g_2 \in K, then g_1g_2 \in K. PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]