Cyclic subgroup

All prime-ordered groups are cyclic

If a group 𝐺 has prime order 𝑝 =|𝐺|, then it must be cyclic, #m/thm/group i.e. isomorphic to +𝑝.

Proof

Let 𝐺 be a group of prime order 𝑝. By Lagrange's theorem, 𝐺 only has subgroups of order 1 and 𝑝. Since 𝑝 >1, there exists 𝑎 𝐺 such that 𝑎 𝑒. Then |𝑎| =𝑝 and therefore 𝑎 =𝐺.

Clearly +𝑝 is a simple group.

See also


#state/tidy | #lang/en | #SemBr