TheoremBase

Left Cosets Partition a Group and All Have the Same Cardinality

Statement

Let (G,∗)(G,\ast) be a group, written multiplicatively, with identity element eGe_G and inverses a−1a^{-1} as in Uniqueness of the Identity Element and of Inverses in a Group, and let HH be a subgroup of GG. For a∈Ga\in G let aHaH be the left coset determined by aa. Then for all a,b∈Ga,b\in G the following hold.

  1. a∈aHa\in aH. In particular every element of GG lies in at least one left coset, and H=eGHH=e_GH is itself a left coset.
  2. aH=bHaH=bH if and only if a−1b∈Ha^{-1}b\in H.
  3. If aH∩bH≠∅aH\cap bH\ne\emptyset, then aH=bHaH=bH. Consequently every element of GG lies in exactly one left coset of HH.
  4. The map h↦ahh\mapsto ah is a bijection from HH onto aHaH. In particular, if HH has tt elements for some natural number tt, then every left coset aHaH has tt elements.

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