Theorem-G satises the right and left cancellation laws
Let G be a group. (i) G satises the right and left cancellation laws; that is, if a; b; x ≡ G, then ax = bx and xa = xb each imply that a = b. (ii) If g ≡ G, then (g-1)-1 = g.
Let G be a group.
(i) G satises the right and left cancellation laws; that is, if a; b; x ≡ G, then ax = bx and xa = xb each imply that a = b.
(ii) If g ≡ G, then (g-1)-1 = g.
Expert
Proof:
(i) From ax = bx, we have axx-1 = bxx-1, then ae = be, then a = b. Similarly for the other case.(ii) Temporarily denote the inverse of g-1 by h (instead of (g-1)-1). Then the defining property of h, from the axiom for inverses applied to g-1, is that
g-1h = hg-1 = e:
But g itself satises these equations in place of h, because the axiom for inverses applied to g says that
gg-1 = g-1g = e:
Hence, since inverses are unique, h = (g-1)-1 = g, as required.
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Let G be a group. (i) G satises the right and left cancellation laws; that is, if a; b; x ≡ G, then ax = bx and xa = xb each imply that a = b. (ii) If g ≡ G, then (g-1)
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