# 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 $$\mathbb{F}$$?

• This is a really nice question. I think I might have a partial answer. If $K$ and $R$ are both algebraic groups (and hence Zariski closed subgroups), this semidirect product is linear because it is an affine algebraic group and all such groups are linear. The reason we still get an algebraic group is because we can do everything in terms of Hopf algebras, where a similar notion of semi-direct product exists. So in particular, the semidirect product of two $GL_{n}$'s is linear. This isn't much of an answer though. – Siddharth Venkatesh Dec 5 '14 at 8:31
• @SiddharthVenkatesh Unfortunately my knowledge of algebraic groups and Hopf algebras is rather poor. Could you give me some references so that I can have a look? – Michal Ferov Dec 6 '14 at 13:27
• Sure. It would be helpful if you could let me know about your background in representation theory and algebraic geometry though because the books I have in mind have different prerequisites. – Siddharth Venkatesh Dec 6 '14 at 21:49
• I have reasonable understanding of algebraic geometry, mainly in positive characteristic, representation theory is bit weaker, just the basics. – Michal Ferov Dec 7 '14 at 22:49
• Ok. Try looking up James Humphreys' "Linear Algebraic Groups". Section 8.6 in particular proves that every affine algebraic group is linear. Humphreys does this only in algebraically closed fields but it holds in general (a reference here would be Milne's notes on Algebraic Groups.) I don't think you need to know too much representation theory to read the former but you might need more background for Milne's notes. I don't know a reference for the Hopf algebra stuff unfortunately but I can explain what I was talking about if you know the basic definitions of a Hopf algebra. – Siddharth Venkatesh Dec 7 '14 at 23:17

$$\def\ZZ{\mathbb{Z}}\def\QQ{\mathbb{Q}}\def\CC{\mathbb{C}}$$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. $$\square$$