Suppose x, y G. Since x and y have order 1 or 2, we have x^-1 = x and y^-1 = y. Consequently,

x^-1y^-1xy = xyxy = (xy)² = e.

Multiplying the extreme sides by yx from the left, we find xy = yx.

Let q = p^a be the highest power of p occurring in |G|, so |G| = qm with gcd(q,m) = 1.

If m = 1, take an element x in G\{e}. By Lagrange's theorem, x has order a divisor of q, and hence a power of p, say p^b. Then x^pb-1 has order p. Hence the result for m = 1.

As a consequence, Part 2 holds for |G| < 6. We proceed by induction on |G|.

So, assume the truth of the assertion for all groups of order smaller than |G|. Consider the set X_q of all subsets of G of size q. The group G acts on X_p by left multiplication: the element g G carries {a₁, ..., a_q} to {ga₁, ..., ga_q}. Observe that, by a binomial argument

|Xq| =

qm q

0 mod p.

Hence, there is an orbit of G on X_q of size not divisible by p. According to a previous theorem, an element, say Y, of this orbit in X_q has a stabilizer, say H, in G of order divisible by q. On the other hand, H cannot be all of G, as left multiplication is transitive on G, and so left multiplication by G does not leave invariant the subset Y of size q. Hence the induction hypothesis applies to H, yielding that it contains an element of order p; but then so does G.