Let G be a group and let H ≤ G. Suppose that no other subgroup of G has the same cardinality as H. Prove that H ⊴ G.
Added by Josep H.
Step 1
e., for any g ∈ G, we have gHg⁻¹ ⊆ H. Suppose, for the sake of contradiction, that there exists some g ∈ G such that gHg⁻¹ ∩ H = ∅. Then, consider the subgroup K = ⟨H, gHg⁻¹⟩ generated by H and gHg⁻¹. Since H and gHg⁻¹ are both subgroups of G, K is also a Show more…
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