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For the classical sequence spaces $\ell_p$ ($p\not=2$) and $c_0$ each surjective linear isometry $U$ has the form $U(a_i)=(\varepsilon_i a_{\pi(i)})$ for a permutation $\pi$ of $\mathbb{N}$ and $\varepsilon_i \in\{\pm 1\}$. Spaces that have the property are said to have a standard group of isometries.

My basic question is: What natural group is the above group isomorphic to. Unless I have made a mistake, it does not seem to be isomorphic to $S_{\infty} \times \Pi_{i=1}^\infty \mathbb{Z}_2$.

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    $\begingroup$ Isn't it $( \mathbb{Z}/2\mathbb{Z}) \wr {\rm Sym}(\mathbb{N})?$ $\endgroup$
    – Geoff Robinson
    Commented Sep 3, 2021 at 18:04
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    $\begingroup$ It should be the wreath product $\mathbb Z_2\wr S_\infty\cong\prod\mathbb Z_2\rtimes S_\infty$ $\endgroup$
    – Wojowu
    Commented Sep 3, 2021 at 18:05
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    $\begingroup$ You mean the real-valued $\ell^p$, as pointed out in your previous post. To clarify @Wojowu's comment, and also replacing $N$ with an arbitrary set $X$ (I always find awkward, albeit widely done in some communities, to write $S_\infty$ as if uncountable sets were not existing), it's the unrestricted permutational wreath product $(Z/2Z)\wr Sym(X)=(Z/2Z)^X\rtimes Sym(X)$. In the complex case, it's also the same wreath product, replacing $Z/2Z$ with the circle group. $\endgroup$
    – YCor
    Commented Sep 3, 2021 at 18:07
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    $\begingroup$ @Wojowu and I seem to be saying the same thing, only in slightly different notation. $\endgroup$
    – Geoff Robinson
    Commented Sep 3, 2021 at 18:07
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    $\begingroup$ That group is sometimes called the infinite hyperoctahedral group. $\endgroup$
    – Tomasz Kania
    Commented Sep 3, 2021 at 18:08

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