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For (1), the answer for finite permutations (as defined by the OP) is clearly yes. This of course is a characterization of finite permutations. A permutation is finite if and only if it is the product of a finite number of cycles.

For non-finite permutations, we'll have to use an infinite product of cycles or transpositions to make sense of (1). So we'll need to define what we mean by this. A reasonable definition of an infinite product of cycles is that the product should be well-defined if we read it from right to left.

Let me explain what this means. By reading an infinite product from right to left we may regard the set of cycles in the product to have a certain order type $\alpha$. Note that $\alpha = \prod_i^\beta \alpha_i$, where each $\alpha_i \leq \omega$ and $\beta \leq \omega$.

So, for each $\alpha_i$ in $\alpha$ and for each $n \in \mathbb{N}$ we first insist that the sequence $f_1(n), f_2(n), \dots $ is eventually constant where $f_i$ is the product of the first $i$ cycles which appear in $\alpha_i$. By convention this is satisfied if $\alpha_i$ is finite. With this condition, each $\alpha_i$ induces a function $g_i: \mathbb{N} \to \mathbb{N}$. The final condition is that for each $n$ the sequence $g_1(n), g_2g_1(n), g_3g_2g_1(n), \dots, $ is eventually constant.

With this definition the permutation $\sigma$ given by Michael can be written as $...(78)(56)(34)(12)$ which is well-defined, with $\alpha=\omega$.

For (1), the answer for finite permutations (as defined by the OP) is clearly yes. This of course is a characterization of finite permutations. A permutation is finite if and only if it is the product of a finite number of cycles.

For non-finite permutations, we'll have to use an infinite product of cycles or transpositions to make sense of (1). So we'll need to define what we mean by this. A reasonable definition of an infinite product of cycles is that the product should be well-defined if we read it from right to left.

Let me explain what this means. By reading an infinite product from right to left we may regard the set of cycles in the product to have a certain order type $\alpha$. Note that $\alpha = \prod_{i=1}^\beta \alpha_i$, where each $\alpha_i \leq \omega$ and $\beta \leq \omega$.

So, for each $\alpha_i$ in $\alpha$ and for each $n \in \mathbb{N}$ we first insist that the sequence $f_1(n), f_2(n), \dots $ is eventually constant where $f_i$ is the product of the first $i$ cycles which appear in $\alpha_i$. By convention this is satisfied if $\alpha_i$ is finite. With this condition, each $\alpha_i$ induces a function $g_i: \mathbb{N} \to \mathbb{N}$. The final condition is that for each $n$ the sequence $g_1(n), g_2g_1(n), g_3g_2g_1(n), \dots, $ is eventually constant.

With this definition the permutation $\sigma$ given by Michael can be written as $...(78)(56)(34)(12)$ which is well-defined, with $\alpha=\omega$. Indeed, Douglas' excellent answer shows that any permutation can be written in this way (with $\alpha=\omega$).

For (1), the answer for finite permutations (as defined by the OP) is clearly yes. This of course is a characterization of finite permutations. A permutation is finite if and only if it is the product of a finite number of cycles.

For non-finite permutations, we'll have to use an infinite product of cycles or transpositions to make sense of (1). So we'll need to define what we mean by this. A reasonable definition of an infinite product of cycles is that the product should be well-defined if we read it from right to left.

Let me explain what this means. By reading an infinite product from right to left we may regard the set of cycles in the product to have a certain order type $\alpha$. Note that $\alpha = \prod_i^\beta \alpha_i$, where each $\alpha_i \leq \omega$ and $\beta \leq \omega$.

So, for each $\alpha_i$ in $\alpha$ and for each $n \in \mathbb{N}$ we first insist that the sequence $f_1(n), f_2(n), \dots $ is eventually constant where $f_i$ is the product of the first $i$ cycles which appear in $\alpha_i$. By convention this is satisfied if $\alpha_i$ is finite. With this condition, each $\alpha_i$ induces a function $g_i: \mathbb{N} \to \mathbb{N}$. The final condition is that for each $n$ the sequence $g_1(n), g_2g_1(n), g_3g_2g_1(n), \dots, $ is eventually constant.

With this definition the permutation $\sigma$ given by Michael can be written as $...(78)(56)(34)(12)$ which is well-defined. Furthermore, Joel's answer shows that any permutation can be written in this way.

For (1), the answer for finite permutations (as defined by the OP) is clearly yes. This of course is a characterization of finite permutations. A permutation is finite if and only if it is the product of a finite number of cycles.

For non-finite permutations, we'll have to use an infinite product of cycles or transpositions to make sense of (1). So we'll need to define what we mean by this. A reasonable definition of an infinite product of cycles is that the product should be well-defined if we read it from right to left.

Let me explain what this means. By reading an infinite product from right to left we may regard the set of cycles in the product to have a certain order type $\alpha$. Note that $\alpha = \prod_i^\beta \alpha_i$, where each $\alpha_i \leq \omega$ and $\beta \leq \omega$.

So, for each $\alpha_i$ in $\alpha$ and for each $n \in \mathbb{N}$ we first insist that the sequence $f_1(n), f_2(n), \dots $ is eventually constant where $f_i$ is the product of the first $i$ cycles which appear in $\alpha_i$. By convention this is satisfied if $\alpha_i$ is finite. With this condition, each $\alpha_i$ induces a function $g_i: \mathbb{N} \to \mathbb{N}$. The final condition is that for each $n$ the sequence $g_1(n), g_2g_1(n), g_3g_2g_1(n), \dots, $ is eventually constant.

With this definition the permutation $\sigma$ given by Michael can be written as $...(78)(56)(34)(12)$ which is well-defined, with $\alpha=\omega$.

For (1), the answer for finite permutations (as defined by the OP) is clearly yes. This of course is a characterization of finite permutations. A permutation is finite if and only if it is the product of a finite number of cycles.

For non-finite permutations, we'll have to use an infinite product of cycles or transpositions to make sense of (1). So we'll need to define what we mean by this. A reasonable definition of an infinite product of cycles is that the product should be well-defined if we read it from right to left.

Let me explain what this means. By reading an infinite product from right to left we may regard the set of cycles in the product to have a certain order type $\alpha$. Note that $\alpha = \prod_i^\beta \alpha_i$, where each $\alpha_i \leq \omega$ and $\beta \leq \omega$.

So, for each $\alpha_i$ in $\alpha$ and for each $n \in \mathbb{N}$ we first insist that the sequence $f_1(n), f_2(n), \dots $ is eventually constant where $f_i$ is the product of the first $i$ cycles which appear in $\alpha_i$. With this condition, each $\alpha_i$ induces a function $g_i: \mathbb{N} \to \mathbb{N}$. The final condition is that for each $n$ the sequence $g_1(n), g_2g_1(n), g_3g_2g_1(n), \dots, $ is eventually constant.

With this definition the permutation $\sigma$ given by Michael can be written as $...(78)(56)(34)(12)$ which is well-defined. Furthermore, Joel's answer shows that any permutation can be written in this way.

For (1), the answer for finite permutations (as defined by the OP) is clearly yes. This of course is a characterization of finite permutations. A permutation is finite if and only if it is the product of a finite number of cycles.

For non-finite permutations, we'll have to use an infinite product of cycles or transpositions to make sense of (1). So we'll need to define what we mean by this. A reasonable definition of an infinite product of cycles is that the product should be well-defined if we read it from right to left.

Let me explain what this means. By reading an infinite product from right to left we may regard the set of cycles in the product to have a certain order type $\alpha$. Note that $\alpha = \prod_i^\beta \alpha_i$, where each $\alpha_i \leq \omega$ and $\beta \leq \omega$.

So, for each $\alpha_i$ in $\alpha$ and for each $n \in \mathbb{N}$ we first insist that the sequence $f_1(n), f_2(n), \dots $ is eventually constant where $f_i$ is the product of the first $i$ cycles which appear in $\alpha_i$. By convention this is satisfied if $\alpha_i$ is finite. With this condition, each $\alpha_i$ induces a function $g_i: \mathbb{N} \to \mathbb{N}$. The final condition is that for each $n$ the sequence $g_1(n), g_2g_1(n), g_3g_2g_1(n), \dots, $ is eventually constant.

With this definition the permutation $\sigma$ given by Michael can be written as $...(78)(56)(34)(12)$ which is well-defined. Furthermore, Joel's answer shows that any permutation can be written in this way.