Past exam of the mathematics course of the University of Oxford / 2014 / Prelims / Mathematics I / 5 / a / i

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- Solution i

Solution

Among the

∣ G ∣ + 1

e, g, g^{2}, \dots, g^{∣ G ∣}

in the finite group

G

, the pigeonhole principle forces two to be equal:

g^{i} = g^{j}

with

0 \leq i < j \leq ∣ G ∣

. Multiply by

g^{- i}

to obtain

g^{j - i} = e, j - i > 0.

This proves the existence of a positive power equal to the identity element, without presupposing the order of a group element.

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