# Generating an infinite group with more than one commutative generator of finite order

Is it possible to generate an infinite group if all of the groups generators are finite order and commute?

In the simplest case, let $$G$$ be a group and $$a,b \in G$$, $$aa = bb = e$$ ($$e$$ is the identity), $$ab = ba$$, $$\langle a,b \rangle = G$$. Because $$a$$ and $$b$$ commute, any combination of them could be transposed and ultimately result in $$a$$, $$b$$, or $$e$$. For example, $$a^{-1}ba = a^{-1}ab = eb = b$$.

If the generators do not commute, however, an infinite group may be possible, but commuting generators of finite order guarantee a finite group.

Is this reasoning correct?

• Yes, your reasoning is correct but incomplete. A careful proof that commuting generators imply commutativity of a group, is by induction on the word length of the elements of the given group. – Moishe Kohan Dec 6 '19 at 13:16

It is true that a group which is generated by finitely many commuting elements of finite order has finite order. Basically, it is a matter of proving that the function $$f:\langle a_1\rangle\oplus\cdots\oplus \langle a_n\rangle\to G\\ f(a_1^{m_1},\cdots,a_n^{m_n})=a_1^{m_1}\cdots a_n^{m_n}$$ is a surjective group homomorphism.
Your guess on the non-commutative case is correct, see for instance the group $$\Bbb Z\rtimes\Bbb F_2$$, $$(n,x)(m,y)=(n+(-1)^xm,x+y)$$, which is generated by its subset $$\{(3,1),(2,1)\}$$.
However, it is possible for an abelian group generated by elements of finite order not to have infinite order when it isn't finitely generated. For instance, the group $$\Bbb F_2^{\Bbb N}$$, the additive group of functions $$\Bbb N\to \Bbb F_2$$, has the cardinality of $$\Bbb R$$ and all its elements satisfy $$f+f=0$$.