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I have a question for the following proof.

Let G = {$f_n : n \epsilon \mathbb{Z}$} prove that G is a subgroup of $S_\mathbb R$

with $f_n$ defined by for each n $f_n$(x) = x + n.

If S is a symmetric group, how is G a subgroup of S? Wouldn't G contain all integers, while S would be a limited set? I understand to prove that something is a subgroup I have to show that it is not empty, closed under the same operation, and closed under inverses.

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  • $\begingroup$ The elements of $G$ are functions from $\mathbb{R}$ to $\mathbb{R}$, not integers. Also, the operation is composition of functions.. $\endgroup$
    – quasi
    Apr 26, 2017 at 23:18
  • $\begingroup$ Hints: $\;(1)\;$Show each $f_n$ is a bijection from $\mathbb{R}$ to $\mathbb{R}$;$\;(2)\;$How can you express the result of the composition $(f_m \circ f_n)(x)$?;$\;(3)\;$What is the inverse of $f_n$? $\endgroup$
    – quasi
    Apr 26, 2017 at 23:24
  • $\begingroup$ I think I understand that $f_n$ is shifting each element in $\mathbb{R}$ by some number, so it's just a permutation. What does showing that its a bijection do? $\endgroup$
    – user406955
    Apr 26, 2017 at 23:33
  • $\begingroup$ By defintition, $S_\mathbb{R}$ is the set of bijections from $\mathbb{R}$ to $\mathbb{R}$. $\endgroup$
    – quasi
    Apr 26, 2017 at 23:37

1 Answer 1

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By definition, the symmetric group $S_\mathbb{R}$ is the set of all bijections from $\mathbb{R}$ to $\mathbb{R}$.

For each $n \in \mathbb{Z}$, let $f_n\colon\mathbb{R}\rightarrow\mathbb{R}$ be defined by $f_n(x) = x + n$.

Let $G = \{f_n \mid n \in \mathbb{Z}\}$.

Outline:

  • Note that $f_0$ is the identity map on $\mathbb{R}$.
  • Show that $f_m \circ f_n = f_{m+n}$.
  • Deduce that $G$ is closed under composition.
  • What is the result of $f_n \circ f_{-n}$ and $f_{-n} \circ f_{n}$?
  • Deduce that each $f_n$ is bijective.
  • In addition, you get that each $f_n$ has an inverse in $G$.
  • Conclude that $G$ is a subgroup of $S_\mathbb{R}$.
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  • $\begingroup$ I can see that G is closed under function composition now, but is function composition the same operation for $S_R$? $\endgroup$
    – user406955
    Apr 27, 2017 at 0:25
  • $\begingroup$ @user406955 -- Yes, by definition, for any set $X$, $S_X$ is the set of bijections from $X$ to $X$, with group operation as composition of functions. $\endgroup$
    – quasi
    Apr 29, 2017 at 22:33

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