Can we create, for any two given natural numbers $m,n$ , a group $G$ having two elements $a,b$ with $|a|=m, |b|=n$ and $|ab|=\infty$

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Can we create, for any two given natural numbers $m,n$ , a group $G$ having two elements $a,b$ with $|a|=m,|b|=n$ and $|ab|=\infty$

Through an exercise in Gallian, I learnt that we can do this in General Linear Group of 2x2 matrices(Feel free to ask for pictures in comments) for elements of order 3 and 4.

Extending this with a basic application of external direct products(With multiplicative group of complex numbers, for instance) , we can create elements of orders $3k_1$ and $4k_2$ $\forall k_1,k_2 \in \Bbb N $ which give their product to have infinite order.

But can this be generalized, for all pairs of natural numbers?

Answer 1

Yes - we can build such a group via generators and relations.

Specifically, the simplest such group is $$G=\langle a, b: a^m=b^n=e\rangle.$$ This is the free group on two generators, modulo the normal subgroup generated by $a^m$ and $b^n$; it's easy to check that $a$ has order $m$, $b$ has order $n$, but $ab$ has infinite order.

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Here's a concrete description of $G$. An element of the free group on two generators is an equivalence class of words in $\{a, b, a^{-1}, b^{-1}\}$; the equivalence relation is, "can be gotten from each other by adding/removing strings of the form "$aa^{-1}$," "$a^{-1}a$", "$bb^{-1}$," and "$b^{-1}b$". For example, $$\mbox{"$abaa$," "$abaa^{-1}b^{-1}baa$," and $"abaabb^{1}$"}$$ each represent the same element.

The quotient group $G$ is defined the same way, but we have a broader class of "deletable/addable" strings: namely, in addition to the four above, we also have "$a^m$" and "$b^n$".

Answer 2

There are groups in which for all pairs of natural numbers $m,n>1$ there are elements of orders $m$ and $n$ with product of infinite order.

E.g. in the group of permutations of $\mathbb{N}$ take, for any $m,n>1$:

$g=(0, ... ,m-1)(m+n-2, ... ,2m+n-3)(2m+2n-4, ... ,3m+2n-5) ...$

and

$h=(m-1, ... ,m+n-2)(2m+n-3, ... ,2m+2n-4)$ $\ \ \ \ \ \ \ \ \ \ \ \ (3m+2n-5, ... ,3m+3n-6) ...$

Then $g$ has order $m$, $h$ has order $n$ and $gh$ has infinite order.

Was that what you were asking?

P.S. By $gh$ here I mean first apply $g$, then apply $h$. Many people do this back to front. But whichever way, you still get an element of infinite order because $o(hg)=o(h^{-1}(hg)h)=o(gh)$. It's just more awkward to calculate.