Let $G$ denote the group of bijective maps $g : \mathbb{Z}\to \mathbb{Z}$ such that $g$ fixes all but finitely many integers. Show that there does not exist a field $F$ and an $n\ge 1$ so that $G$ is isomorphic to a subgroup of (embeds in) $GL_n(F),$ the set of invertible $n\times n$ matrices with entries in $F$.
I know that for any field $F$, every finite group embeds into $GL_n(F)$ for some $n\ge 1.$ I think it might be useful to use a contradiction here. So suppose there exists an $n\ge 1$ and a field $F$ so that $\phi : G\to S$ is an isomorphism, where $S$ is a subgroup of $GL_n(F),$ denoted $S\leq GL_n(F).$ There are various properties of isomorphisms that may be useful; for instance, the order of $g\in G$ equals the order of $\phi(g)$ in $S$, $S$ is abelian if and only if $G$ is abelian, etc. I'm not sure how to use the properties of $G$ and $\phi$ to obtain a contradiction here though.
Edit: I was wondering if @JyrkiLahtonen could elaborate on his answer? I think I mostly understand it, but I don't get a few details. Below is my understanding of his answer.
$G_2$ is abelian because the permutations $\sigma_i$ and $\sigma_j$ commute for any $i$ and $j$ and any element of $G_2$ is of the form $\sigma_{i_1}^{a_{i_1}}\cdots \sigma_{i_n}^{a_{i_n}}$ for some $i_j \ge 1, a_{i_j} \in \mathbb{Z}\,\forall i_j$ (so $\sigma_{a}^{b}\sigma_{c}^{d} = \sigma_{c}^{d}\sigma_{a}^{b}$ for any $b,d \in\mathbb{Z}, a,c\ge 1$). It is an infinite direct sum of cyclic groups of order two because for any $i$ the cyclic group $\langle \sigma_i \rangle = \{e, \sigma_i\}$ has trivial intersection with the subgroup $\langle \{ \langle \sigma_j\rangle : j\neq i\}\rangle$, $G_2$ is abelian so every subgroup $\langle \sigma_i\rangle$ of $G_2$ is normal in $G_2$, and $G_2 = \langle \{ \langle \sigma_j\rangle : j\ge 1\} \rangle = \langle e,\sigma_1,\sigma_2,\cdots \rangle = \langle \sigma_1,\sigma_2,\cdots \rangle.$
The vectors $\frac{1}2(x+\phi(\sigma_i)(x))$ and $\frac{1}2(x-\phi(\sigma_i)(x))$ are eigenvectors of $\phi(\sigma_i)$ with corresponding eigenvalues +1 and $-1$. Since $x$ was arbitrary, this shows that every vector of $V$ is a linear combination of eigenvectors of $\phi(\sigma_i),$ so the set of eigenvectors of $\phi(\sigma_i)$ spans $V$. Hence we can obtain a basis of eigenvectors of $\phi(\sigma_i)$ for $V$ (e.g. we start with an eigenvector $v_1$ and then if $V\neq \mathrm{span} \{v_1\}$, we pick an eigenvector $v_2 \in V\neq \mathrm{span} \{v_1\}$ and continue until we get a basis). Since $V$ has an ordered basis consisting of eigenvectors of $\phi(\sigma_i)$, the matrix $\phi(\sigma_i)$ is diagonalizable.
$N$ is the number of matrices. The above shows that the space $V$ has a basis consisting of eigenvectors of $\phi(\sigma_i)$ for fixed $i$. Every element of $V$ can be written as a linear combination of two eigenvectors of $\phi(\sigma_{k+1})$ corresponding to the eigenvalues $+1$ and $-1$. Also, these eigenspaces are disjoint because if $0\neq w\in V_+\cap V_-,$ then $w$ is an eigenvector of $\phi(\sigma_{k+1})$ with eigenvalue $1$ and $-1$, which isn't possible. So $V= V_+\oplus V_-.$
The transformations $\phi(\sigma_j), 1\leq j\leq k$ commute with $\phi(\sigma_{k+1})$ because $\phi(\sigma_j)\phi(\sigma_{k+1}) = \phi(\sigma_j \sigma_{k+1}) = \phi(\sigma_{k+1}\sigma_j) = \phi(\sigma_{k+1})\phi(\sigma_j).$ Fix $1\leq j\leq k.$ Let $v_1$ and $v_2$ be the eigenvectors of $\phi(\sigma_{k+1})$ corresponding to eigenvalues $1$ and $-1$. Then $\phi(\sigma_j)(v_1) = \phi(\sigma_j)(\phi(\sigma_{k+1})(v_1)) = \phi(\sigma_{k+1})(\phi(\sigma_j)(v_1)),$ so $\phi(\sigma_j)(v_1)$ is an eigenvector of $\phi(\sigma_{k+1})$ with eigenvalue $1.$ Hence $\phi(\sigma_j)(V_+) \subseteq V_+.$ Similarly $\phi(\sigma_j)^{-1}(V_+)\subseteq V_+$ so $V_+ = \phi(\sigma_j) (V_+).$ Similarly $\phi(\sigma_j)(V_-) = V_-.$
So does the induction hypothesis apply for $N=k$?
An eigenvector $v$ is a shared eigenvector of all $\phi(\sigma_i)$ if it's an eigenvector of every $\phi(\sigma_i),$ right?
What does "they must all respect the decomposition (*)" mean precisely?
Hence if $v$ is a shared eigenvector of all $\phi(\sigma_j)$'s in the basis for $V_-,$ for any $1\leq j\leq k, \phi(\sigma_{k+1})(v) = \phi(\sigma_{k+1})(-\phi(\sigma_j)(v)) = -\phi(\sigma_j)(\phi(\sigma_{k+1})(v))$, so $\phi(\sigma_{k+1})(v)$ is an eigenvector of $\phi(\sigma_j)$ with eigenvalue -1. But is $v$ an eigenvector of $\phi(\sigma_{k+1})$ with eigenvalue $-1$?
How does the claim follow from the fact that $GL_n(F)$ contains only $2^n$ diagonal matrices with entries $\pm 1$?
The statement is clearly true, but each $\phi(\sigma_j)$ is diagonalizable, so there is an invertible matrix $P_j$ so that $P_j^{-1} \phi(\sigma_j) P_j$ is diagonal for all $j$.
Why can we adjoin a primitive third root of unity, $\omega$ to the field $F$? Does this just mean we replace $F$ with $F\cup \{\omega\}$?
Best Answer
Hint: The group $G$ has a lot of commuting elements, since any disjoint cycles commute. On the other hand, it's relatively difficult for matrices to commute. For instance, when diagonalizable matrices all commute with each other, they are simultaneously diagonalizable, which severely restricts their possible properties. See if you can use this to get a contradiction.
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