Question

Prove or disprove: If G is a finite group and some element of G has order equal

to the size of G, then G is cyclic.

Answer 1

Solution:

Proof;

Let g$\epsilon$ G have an order n=$\#$ (G).

For each i with 1$\leq$ i$<$ n we have gⁱ$\ne$ e,the identity of G.

The claim herein is that G=$\lang g\rang$

For this ,there are n elements of {gⁱ:0$\le$ i<n}

If gⁱ=g^j for some j<i, multiply both sides of the equation by g^-i=g^i(-1)

We have, e=g^j-i even though 1$\leq$ j-i$<$ n

Contrary to the above observations.

Hence G=$\lang g\rang$ is cyclic.