Is a semidirect product of linear groups a linear group? 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 $\mathbb{F}$?
 A: $\def\ZZ{\mathbb{Z}}\def\QQ{\mathbb{Q}}\def\CC{\mathbb{C}}$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. $\square$
