Of course there are commutative groups with elements that do not have order 2. Indeed, any cyclic group is commutative.

Part 1 does not hold if 2 is replaced by 3 (or a larger prime). There is a non-commutative group of order 27 all of whose elements have order 3.

Part 2 of the lemma is a special case of a more general result. If q is the highest power of a prime p dividing |G|, then G has a subgroup of order q. Such a group is known as a Sylow subgroup of G. However, it is not true that, for each divisor m of |G|, there is a subgroup of G of order m.

For instance, if G = A₅, there is no subgroup of order 30. (Proof: such a subgroup H has index 2 in G and so is a normal subgroup. On the other hand, it contains an element a of order 3; but all elements of order 3 in A₅ are 3-cycles and conjugate to a power of a in G. So normality of H implies that all elements of order 3 are contained in H. But in Chapter 5 we allready noticed that the 3-cycles generate all of G.)