Proof 1

We first prove the special case that where G is abelian, and then the general case; both proofs are by induction on n = |G|, and have as starting case n = p which is trivial because any non-identity element now has order p. Suppose first that G is abelian. Take any non-identity element a, and let H be the cyclic group it generates. If p divides |H|, then a^|H|/p is an element of order p. If p does not divide |H|, then it divides the order [G:H] of the quotient group G/H, which therefore contains an element of order p by the inductive hypothesis. That element is a class xH for some x in G, and if m is the order of x in G, then x^m = e in G gives (xH)^m = eH in G/H, so p divides m; as before x^m/p is now an element of order p in G, completing the proof for the abelian case.

In the general case, let Z be the center of G, which is an abelian subgroup. If p divides |Z|, then Z contains an element of order p by the case of abelian groups, and this element works for G as well. So we may assume that p does not divide the order of Z. Since p does divide |G|, and G is the disjoint union of Z and of the conjugacy classes of non-central elements, there exists a conjugacy class of a non-central element a whose size is not divisible by p. But the class equation shows that size is [G : C_G(a)], so p divides the order of the centralizer C_G(a) of a in G, which is a proper subgroup because a is not central. This subgroup contains an element of order p by the inductive hypothesis, and we are done.

Proof 2

This proof uses the fact that for any action of a (cyclic) group of prime order p, the only possible orbit sizes are 1 and p, which is immediate from the orbit stabilizer theorem.

The set that our cyclic group shall act on is the set

${\displaystyle X=\{\,(x_{1},\ldots ,x_{p})\in G^{p}:x_{1}x_{2}\cdots x_{p}=e\,\}}$

of p-tuples of elements of G whose product (in order) gives the identity. Such a p-tuple is uniquely determined by all its components except the last one, as the last element must be the inverse of the product of those preceding elements. One also sees that those p − 1 elements can be chosen freely, so X has |G|^p−1 elements, which is divisible by p.

Now from the fact that in a group if ab = e then also ba = e, it follows that any cyclic permutation of the components of an element of X again gives an element of X. Therefore one can define an action of the cyclic group C_p of order p on X by cyclic permutations of components, in other words in which a chosen generator of C_p sends

${\displaystyle (x_{1},x_{2},\ldots ,x_{p})\mapsto (x_{2},\ldots ,x_{p},x_{1})}$.

As remarked, orbits in X under this action either have size 1 or size p. The former happens precisely for those tuples ${\displaystyle (x,x,\ldots ,x)}$ for which ${\displaystyle x^{p}=e}$. Counting the elements of X by orbits, and reducing modulo p, one sees that the number of elements satisfying ${\displaystyle x^{p}=e}$ is divisible by p. But x = e is one such element, so there must be at least p − 1 other solutions for x, and these solutions are elements of order p. This completes the proof.

Example 1

Let G be a finite group where x² = e for all elements x of G. Then G has order 2ⁿ for some non negative integer n.

To show this, let |G| equal m. If m is 1, then G = {e} and the result is valid for ${\displaystyle n=0}$. In the case of m ≥ 2, assume by contradiction that m has an odd prime factor p. Then, by Cauchy's theorem, G has an element x with order p. As ${\displaystyle p>2}$, this implies that ${\displaystyle x^{2}\neq e}$, which conflicts with the initial hypothesis. Therefore m must be 2ⁿ.^[5]

It can be shown that G is also an abelian group. A group that satisfies the property of this example is called an elementary abelian 2-group or Boolean group. The well-known example is Klein four-group.

Example 2

An abelian simple group is either {e} or cyclic group C_p whose order is a prime number p. Let G be an abelian group, then all subgroups of G are normal subgroups. So, if G is a simple group, G has only normal subgroup that is either {e} or G. If |G| = 1, then G is {e}. It is suitable. If |G| ≥ 2, let a ∈ G is not e, the cyclic group ${\displaystyle \langle a\rangle }$ is subgroup of G and ${\displaystyle \langle a\rangle }$ is not {e}, then ${\displaystyle G=\langle a\rangle .}$ Let n is the order of ${\displaystyle \langle a\rangle }$. If n is infinite, then

${\displaystyle G=\langle a\rangle \supsetneqq \langle a^{2}\rangle \supsetneqq \{e\}.}$

So in this case, it is not suitable. Then n is finite. If n is composite, n is divisible by prime q which is less than n. From Cauchy's theorem, the subgroup H will be exist whose order is q, it is not suitable. Therefore, n must be a prime number.