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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?

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    $\begingroup$ 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. $\endgroup$ – Moishe Kohan Dec 6 '19 at 13:16
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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 $f+f=0$.

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