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It is known that linear groups are not closed under extensions, but what if the extension splits, i.e. it is a semidirect product?

Suppose that $K,R$ are subgroups of $mathop{GL}(n,mathbb{F})$, where $mathbb{F}$ is a field, and suppose that I have a homomorphism $phi colon R to mathop{Aut}(K)$ which defines the semidirect product $G = K rtimes_{phi}R$. My question is, does $G$ embed into $mathop{GL}(m,mathbb{F})$ for some $m > n$? Or perhaps into $mathop{GL}(m,mathbb{F}')$, where $mathbb{F}'$ is some extension of $mathbb{F}$?

$defZZ{mathbb{Z}}defQQ{mathbb{Q}}defCC{mathbb{C}}$No. Set $p^{-n} ZZ$ to be the set of rational numbers of the form $tfrac{a}{p^n}$ with $a in ZZ$. Put $p^{-infty} ZZ = bigcup_n p^{-n} ZZ$ and let $A$ be the abelian group $(p^{- infty} ZZ)/ZZ$. Then $A$ is linear over $mathbb{C}$, namely, $A$ embeds into $GL_1(CC)$ by $theta mapsto exp(2 pi i theta)$. Let $r$ be an integer which is not divisible by $p$ and not equal to $pm 1$. Let $ZZ$ act on $A$ by multiplication by $r$. The group $ZZ$ is also a subgroup of $GL_1(CC)$. However, we will show that $ZZ ltimes A$ does not embed in $GL_n(CC)$ for any $n$. We write $ZZ_p$ for the $p$-adic integers.

We first work out the representation theory of $A$. For any $p$-adic integer $k$, multiplication by $k$ is a well defined map $A to A$ because, for any $tfrac{a}{p^n}$ in $A$, the product $ka$ is determined by the residue class of $k$ modulo $p^n$. So, for any $k in ZZ_p$, we have a one dimensional representation $chi_k$ of $A$ given by $chi_k(theta) = exp(2 pi i k theta)$. We claim that any finite dimensional representation $V$ of $A$ is a direct sum of these $chi_k$'s.

Proof: If we restrict $V$ to the finite subgroup $p^{-n} ZZ/ZZ$ of $A$, then $V$ decomposes into isotypic components of the form $theta mapsto exp(2 pi i k theta)$ for $k in ZZ/p^n ZZ$. As we increase $n$, isotypic summands may increase, but it can't be more than $dim V$, so it stabilizes for $n$ large. Thus, there is some decomposisition $V = V_1 oplus V_2 oplus cdots oplus V_r$ and, for each $1 leq j leq r$, some sequence $k^j_n in ZZ/p^n ZZ$ such that $p^{-n} ZZ/ZZ$ acts on $V_j$ by $exp(2 pi i k^j_n theta)$. For $m < n$, we have $k^j_n equiv k^j_n bmod p^m$ (because $p^{-m} ZZ/ZZ subset p^{-n} ZZ/ZZ$). So $k^j_n$ approaches some limit $k^j in ZZ_p$, and $A$ acts on $V_j$ by $chi_{k^j}$. $square$

Now we have to show that $ZZ ltimes A$ does not embed in $GL(V)$ for any finite dimensional $CC$ vector space $V$. Suppose otherwise. Restrict $V$ to $A$, so $V$ decomposes as a direct sum of $chi_k$'s, let $K subset ZZ_p$ be the set of $k$'s which occur.

I claim that $K$ is taken to itself under multiplication by $r$. Indeed, if $v in V$ obeys $(0, theta) cdot v = exp(2 pi i k theta) v$, then $(0, theta) cdot (1,0) cdot v = (1,0) cdot (0, r theta) v = (1,0) exp(2 pi i k r theta) v = exp(2 pi i k r theta) (1,0) v$, so multiplication by $(1,0)$ takes $A$-eigenvectors of weight $k$ to $A$-eigenvectors of weight $rk$.

But $|K| leq dim V$, and the only finite orbit of multiplication by $r$ on $ZZ_p$ is ${ 0 }$. So $K = { 0 }$, and the representation $V$ has kernel $A$, contradicting the assumption that the map $(ZZ ltimes A) to GL(V)$ is an embedding. $square$