Lemma·L28
Elementary consequences of the group axioms
One identity, one inverse each, and cancellation on both sides.
In every group
:(i)
;(ii)
, written
;(iii)
and
;(iv)
;(v)
.
In words
A group has only one identity element; each element has only one inverse; equal products with a common factor on the left force the other factors equal, and likewise on the right; inverting twice returns the element; and the inverse of a product is the product of the inverses in reverse order.
Never needed: F03 · F04 · F06 · F08 · F10 · F11 · F12 · A01 · A03 · A04 · A05 · A06 · A07 · A08 (computed from the citation graph, not asserted).
Proof
- 1(i) Let and both be identities (D038 (G2)). Then (using that is an identity, acting on the right) and (using that is an identity, acting on the left). So by transitivity.
- 2(ii) Let and both be inverses of . Then using (G2), the inverse property of , associativity (G1), the inverse property of , and (G2) again (D038, substitutivity).
- 3(iii) Suppose . Multiply on the left by : by (G2), the inverse property, associativity (G1) twice, and the hypothesis. Right cancellation is symmetric, multiplying by on the right.
- 4(iv) The element satisfies and , which says precisely that is an inverse of . By uniqueness (ii), .
- 5(v) Compute, with associativity used freely: and symmetrically . So is an inverse of , and (ii) makes it the inverse.
∎
Remarks
Everything here is two or three lines of equational reasoning from the axioms, but the payoff is notational: the identity, the inverse
, and the freedom to cancel. The socks-and-shoes law (v) (undo in reverse order) and (iv) are exactly what the symmetry and transitivity of the coset relation need in L29.
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