Does a Group Generated by Two Involutions Have to Be Infinite?

A group generated by two elements of order 2 is not necessarily infinite or non-abelian. For example, two distinct commuting involutions generate the Klein four-group, which is finite and abelian. The useful way to determine what happens is to examine the product of the generators.

Reduce the question to one element

Let (G=\langle a,b\rangle), where (a) and (b) have order 2, and define (r=ab). Since (b=ar), the group is also generated by (r) and (a). The involution relations give

$$ara=a(ab)a=ba=(ab)^{-1}=r^{-1}.$$

Thus conjugating (r) by (a) reverses its powers. This relation lets you move every occurrence of (a) in a word to the right. Every element of (G) therefore has one of the forms

$$r^k \quad\text{or}\quad r^k a,$$

for some integer (k). The possible size of the group depends on the order of (r=ab).

When is the group infinite?

If (r) has finite order (n), then its powers reduce to (1,r,\ldots,r^{n-1}). The normal forms show that (G) has at most (2n) elements. In fact, it has exactly (2n): the sets (\langle r\rangle) and (\langle r\rangle a) are disjoint. If they overlapped, then (a) would be a power of (r) and would commute with (r). But (ara=r^{-1}), so that would imply (r=r^{-1}) and hence (r^2=1). If (r) has order 1, then (\langle r\rangle={1}), which cannot contain the nonidentity element (a). If (r) has order 2, an overlap would imply (a=r), and then (r=ab) would give (b=1), also impossible. Thus the two sets are disjoint.

If (r) has infinite order, all its powers are distinct, so (G) is infinite. Conversely, if (G) is infinite, (r) cannot have finite order because that would leave at most (2n) normal forms. Therefore

$$G\text{ is infinite}\quad\Longleftrightarrow\quad ab\text{ has infinite order}.$$

When (r) has finite order (n), the group is a dihedral group of order (2n) (including the small cases: order 2 when (n=1), and the Klein four-group when (n=2)). When (r) has infinite order, it is the infinite dihedral group.

When is the group abelian?

The generators commute exactly when

$$ab=ba.$$

Since (ba=(ab)^{-1}), this is equivalent to (r=r^{-1}), or (r^2=1). Consequently,

$$G\text{ is non-abelian}\quad\Longleftrightarrow\quad (ab)^2\ne 1.$$

If (ab=1), then (a=b) and (G) has order 2. If (ab) has order 2, then (a) and (b) are distinct and generate the Klein four-group, which is abelian. If (ab) has order greater than 2—or has infinite order—the group is non-abelian.

Examples

Finite and abelian: In (C_2\times C_2), take (a=(1,0)) and (b=(0,1)). Both have order 2, they generate the whole group, and they commute. Their product also has order 2.

Finite and non-abelian: In the permutation group (S_3), take (a=(12)) and (b=(23)). These are both of order 2, and their product is a 3-cycle. They generate (S_3), a non-abelian group of order 6.

Infinite and non-abelian: Consider transformations of the real line (a(x)=-x) and (b(x)=2-x). Each is a reflection, so each has order 2. Their product, with composition read right to left, is (ab(x)=x-2), a translation of infinite order. The generated group is therefore infinite and non-abelian.

Common mistakes

• Assuming two involutions force infinitely many elements. The product may have finite order. For example, in (S_3) above, (ab) has order 3 and the generated group has only 6 elements.
• Assuming two involutions cannot commute. Distinct involutions can commute, as they do in the Klein four-group.
• Confusing the order of a generator with the order of its product. Knowing (a^2=b^2=1) does not determine the order of (ab). That order controls whether the generated group is finite and whether it is abelian.