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Lemma·L30

All cosets have the size of the subgroup

Translation is a bijection: every coset is a perfect copy of H.

For every subgroup and every : with the equinumerosity and the left coset.
In words
For every subgroup H of a group G and every element a of G, the left coset aH is equinumerous with the subgroup H itself: multiplying by a merely relabels the elements.
Never needed: F08 · F10 · F11 · F12 · A03 · A04 · A05 · A07 · A08 (computed from the citation graph, not asserted).

Proof

  1. 1
    Define with domain by : as a set, by Separation, and it is a function since each yields exactly one value (L01), landing in by D040.
  2. 2
    Injective: if , then , and left cancellation (L28 (iii)) gives (D014).
  3. 3
    Surjective: every has the form for some (D040), and then (D015).
  4. 4
    So is a bijection , giving ; symmetry of equinumerosity (L22 (ii)) yields .

Remarks

The whole content of "translation by is invertible", concentrated into one bijection; the inverse translation is by . This uniformity is remarkable: however lopsided a subgroup might look inside its group, every coset is an exact-size copy of it. It is one of the two pillars of Lagrange's theorem, the other being the partition into cosets.

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