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Let $S_\infty$ denote the full symmetric group on countably many points and index the points by $\mathbb{N}$. For any weight (for lack of a better name) function $w:\mathbb{N}\rightarrow\mathbb R^+$, define a subgroup $S_\infty(w)\subset S_\infty$ as follows:

$S_\infty(w) := \{\pi\in S_\infty|\sum_{\pi(k)\neq k} w(k)<\infty \}$

The intuitive idea is that given a weight function $w$ a permutation $\pi\in S_\infty$ is in $S_\infty(w)$ if the sum of the weights of all points moved by $\pi$ is finite.

There are two well-known examples of such subgroups:

$\bullet\;\; S_\infty(1)$ is the group of finitely supported permutations of $\mathbb{N}$ or equivalently the direct limit of the finite symmetric groups.

$\bullet\;\; S_\infty(\frac{1}{n^2})$ is simply $S_\infty$ since the series $\{\frac{1}{n^2}\}$ is convergent.

So in this context, I have been thinking about the group $G := S_\infty(\frac{1}{n})$. It is clear $S_\infty(1)\cong\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\backslash\;\; G$ since $G$ contains many subgroups isomorphic to $S_\infty$ (for example the set of all permutations of the points indexed by $b^i$ for $b$ fixed and $i\in\mathbb{N}$).

On the other hand, $G$ contains a proper subgroup isomorphic to the direct sum of countably many copies of $S_\infty$, so in particular it is easy to see $S_\infty\cong\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\backslash\;\; G$ since in $G$ elements of infinite order always have nontrivial centralizers while in $S_\infty$ this is not always the case.

Question: Aside from the description of $G$ as $S_\infty(\frac{1}{n})$, is there any easy to describe/understand presentation of this group?

Primarily I am torn on whether $G$ is itself isomorphic to a direct sum of countably many copies of $S_\infty$; every time I attempt to describe such an isomorphism, there are elements of $G$ not accounted for.

Let $S_\infty$ denote the full symmetric group on countably many points and index the points by $\mathbb{N}$. For any weight (for lack of a better name) function $w:\mathbb{N}\rightarrow\mathbb R^+$, define a subgroup $S_\infty(w)\subset S_\infty$ as follows:

$$S_\infty(w) := \{\pi\in S_\infty|\sum_{\pi(k)\neq k} w(k)<\infty \}$$

The intuitive idea is that given a weight function $w$ a permutation $\pi\in S_\infty$ is in $S_\infty(w)$ if the sum of the weights of all points moved by $\pi$ is finite.

There are two well-known examples of such subgroups:

• $S_\infty(1)$ is the group of finitely supported permutations of $\mathbb{N}$ or equivalently the direct limit of the finite symmetric groups.

• $S_\infty(\frac{1}{n^2})$ is simply $S_\infty$ since the series $\{\frac{1}{n^2}\}$ is convergent.

So in this context, I have been thinking about the group $G := S_\infty(\frac{1}{n})$. It is clear $S_\infty(1)\not\cong G$ since $G$ contains many subgroups isomorphic to $S_\infty$ (for example the set of all permutations of the points indexed by $b^i$ for $b$ fixed and $i\in\mathbb{N}$).

On the other hand, $G$ contains a proper subgroup isomorphic to the direct sum of countably many copies of $S_\infty$, so in particular it is easy to see $S_\infty\not\cong G$ since in $G$ elements of infinite order always have nontrivial centralizers while in $S_\infty$ this is not always the case.

Question: Aside from the description of $G$ as $S_\infty(\frac{1}{n})$, is there any easy to describe/understand presentation of this group?

Primarily I am torn on whether $G$ is itself isomorphic to a direct sum of countably many copies of $S_\infty$; every time I attempt to describe such an isomorphism, there are elements of $G$ not accounted for.

Let $S_\infty$ denote the full symmetric group on countably many points and index the points by $\mathbb{N}$. For any weight (for lack of a better name) function $w:\mathbb{N}\rightarrow\mathbb R^+$, define a subgroup $S_\infty(w)\subset S_\infty$ as follows:

$S_\infty(w) := \{\pi\in S_\infty|\sum_{\pi(k)\neq k} w(k)<\infty \}$

The intuitive idea is that given a weight function $w$ a permutation $\pi\in S_\infty$ is in $S_\infty(w)$ if the sum of the weights of all points moved by $\pi$ is finite.

There are two well-known examples of such subgroups:

$\bullet\;\; S_\infty(1)$ is the group of finitely supported permutations of $\mathbb{N}$ or equivalently the direct limit of the finite symmetric groups.

$\bullet\;\; S_\infty(\frac{1}{n^2})$ is simply $S_\infty$ since the series $\{\frac{1}{n^2}\}$ is convergent.

So in this context, I have been thinking about the group $G := S_\infty(\frac{1}{n})$. It is clear $S_\infty(1)\cong\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\backslash\;\; G$ since $G$ contains many subgroups isomorphic to $S_\infty$ (for example the set of all permutations of the points indexed by $b^i$ for $b$ fixed and $i\in\mathbb{N}$).

On the other hand, $G$ contains a proper subgroup isomorphic to the direct sum of countably many copies of $S_\infty$, so in particular it is easy to see $S_\infty\cong\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\backslash\;\; G$.

Question: Aside from the description of $G$ as $S_\infty(\frac{1}{n})$, is there any easy to describe/understand presentation of this group?

Primarily I am torn on whether $G$ is itself isomorphic to a direct sum of countably many copies of $S_\infty$; every time I attempt to describe such an isomorphism, there are elements of $G$ not accounted for.

Let $S_\infty$ denote the full symmetric group on countably many points and index the points by $\mathbb{N}$. For any weight (for lack of a better name) function $w:\mathbb{N}\rightarrow\mathbb R^+$, define a subgroup $S_\infty(w)\subset S_\infty$ as follows:

$S_\infty(w) := \{\pi\in S_\infty|\sum_{\pi(k)\neq k} w(k)<\infty \}$

The intuitive idea is that given a weight function $w$ a permutation $\pi\in S_\infty$ is in $S_\infty(w)$ if the sum of the weights of all points moved by $\pi$ is finite.

There are two well-known examples of such subgroups:

$\bullet\;\; S_\infty(1)$ is the group of finitely supported permutations of $\mathbb{N}$ or equivalently the direct limit of the finite symmetric groups.

$\bullet\;\; S_\infty(\frac{1}{n^2})$ is simply $S_\infty$ since the series $\{\frac{1}{n^2}\}$ is convergent.

So in this context, I have been thinking about the group $G := S_\infty(\frac{1}{n})$. It is clear $S_\infty(1)\cong\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\backslash\;\; G$ since $G$ contains many subgroups isomorphic to $S_\infty$ (for example the set of all permutations of the points indexed by $b^i$ for $b$ fixed and $i\in\mathbb{N}$).

On the other hand, $G$ contains a proper subgroup isomorphic to the direct sum of countably many copies of $S_\infty$, so in particular it is easy to see $S_\infty\cong\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\backslash\;\; G$ since in $G$ elements of infinite order always have nontrivial centralizers while in $S_\infty$ this is not always the case.

Question: Aside from the description of $G$ as $S_\infty(\frac{1}{n})$, is there any easy to describe/understand presentation of this group?

Primarily I am torn on whether $G$ is itself isomorphic to a direct sum of countably many copies of $S_\infty$; every time I attempt to describe such an isomorphism, there are elements of $G$ not accounted for.

Let $S_\infty$ denote the full symmetric group on countably many points and index the points by $\mathbb{N}$. For any weight (for lack of a better name) function $w:\mathbb{N}\rightarrow\mathbb R^+$, define a subgroup $S_\infty(w)\subset S_\infty$ as follows:

$S_\infty(w) := \{\pi\in S_\infty|\sum_{\pi(k)\neq k} w(k)<\infty \}$

The intuitive idea is that given a weight function $w$ a permutation $\pi\in S_\infty$ is in $S_\infty(w)$ if the sum of the weights of all points moved by $\pi$ is finite.

There are two well-known examples of such subgroups:

$\bullet\;\; S_\infty(1)$ is the group of finitely supported permutations of $\mathbb{N}$ or equivalently the direct limit of the finite symmetric groups.

$\bullet\;\; S_\infty(\frac{1}{n^2})$ is simply $S_\infty$ since the series $\{\frac{1}{n^2}\}$ is convergent.

So in this context, I have been thinking about the group $G := S_\infty(\frac{1}{n})$. It is clear $S_\infty(1)\cong\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\backslash\;\; G$ since $G$ contains many subgroups isomorphic to $S_\infty$ (for example the set of all permutations of the points indexed by $b^i$ for $b$ fixed and $i\in\mathbb{N}$).

On the other hand, $G$ contains a proper subgroup isomorphic to the direct sum of countably many copies of $S_\infty$, so in particular $S_\infty\cong\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\backslash\;\; G$.

Question: Aside from the description of $G$ as $S_\infty(\frac{1}{n})$, is there any easy to describe/understand presentation of this group?

Primarily I am torn on whether $G$ is itself isomorphic to a direct sum of countably many copies of $S_\infty$; every time I attempt to describe such an isomorphism, there are elements of $G$ not accounted for.

Let $S_\infty$ denote the full symmetric group on countably many points and index the points by $\mathbb{N}$. For any weight (for lack of a better name) function $w:\mathbb{N}\rightarrow\mathbb R^+$, define a subgroup $S_\infty(w)\subset S_\infty$ as follows:

$S_\infty(w) := \{\pi\in S_\infty|\sum_{\pi(k)\neq k} w(k)<\infty \}$

The intuitive idea is that given a weight function $w$ a permutation $\pi\in S_\infty$ is in $S_\infty(w)$ if the sum of the weights of all points moved by $\pi$ is finite.

There are two well-known examples of such subgroups:

$\bullet\;\; S_\infty(1)$ is the group of finitely supported permutations of $\mathbb{N}$ or equivalently the direct limit of the finite symmetric groups.

$\bullet\;\; S_\infty(\frac{1}{n^2})$ is simply $S_\infty$ since the series $\{\frac{1}{n^2}\}$ is convergent.

So in this context, I have been thinking about the group $G := S_\infty(\frac{1}{n})$. It is clear $S_\infty(1)\cong\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\backslash\;\; G$ since $G$ contains many subgroups isomorphic to $S_\infty$ (for example the set of all permutations of the points indexed by $b^i$ for $b$ fixed and $i\in\mathbb{N}$).

On the other hand, $G$ contains a proper subgroup isomorphic to the direct sum of countably many copies of $S_\infty$, so in particular it is easy to see $S_\infty\cong\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\backslash\;\; G$.

Question: Aside from the description of $G$ as $S_\infty(\frac{1}{n})$, is there any easy to describe/understand presentation of this group?

Primarily I am torn on whether $G$ is itself isomorphic to a direct sum of countably many copies of $S_\infty$; every time I attempt to describe such an isomorphism, there are elements of $G$ not accounted for.