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Lemma·L32

Laws of powers in a group

Adding exponents multiplies powers, multiplying exponents iterates powers, and the identity is fixed by every power.

Let be a group, , and . With powers from D043: (i) ; (ii) ; (iii) and .
In words
Adding exponents corresponds to multiplying powers; multiplying exponents corresponds to taking a power of a power; the first power of an element is the element itself; and every power of the identity is the identity.
Rests onF02F09F14A01A02
Never needed: F03 · F04 · F05 · F06 · F08 · F10 · F11 · F12 · F13 · A03 · A04 · A05 · A06 · A07 · A08 (computed from the citation graph, not asserted).

Proof

  1. 1
    (i) Fix and induct on (T05). Base : , using (D027), the identity law D038 (G2), and (D043).
  2. 2
    Step: assume . Then using (D027), D043, the hypothesis, associativity D038 (G1), and D043 again. T05 gives (i) for all .
  3. 3
    (iii) First by D043 and D038 (G2) (here ). Next by induction on : (D043), and using D043, the hypothesis, and (G2).
  4. 4
    (ii) Fix and induct on . Base : by D043 and (D028).
  5. 5
    Step: assume . Then using D043, the hypothesis, part (i), and (D028). T05 gives (ii).
∎

Remarks

These are the natural-exponent fragment of the usual index laws. From (i) and commutativity of addition the powers of one element commute with each other, , so every cyclic subgroup is abelian. Full index laws with negative exponents wait on the integers.
Wikipedia ↗pow_add ↗

Used by

L33L35L36T16
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