Diagonalizable linear algebraic group is isomorphic to $(\mathbb{C}^*)^r\times A$, for some finite abelian group $A$ I have three questions about algebraic groups. 
Let $D$ be a linear algebraic group. Then the following are equivalent: 


*

*$D$ is diagonalizable.

*$\mathop{Hom}(D,\mathbb{C}^*)$ is finitely generated with an isomorphism of coordinate rings $\mathbb{C}[D]\cong \mathbb{C}[\mathop{Hom}(D,\mathbb{C}^*)]$. 

*Every rational representation of $D$ is isomorphic to a direct sum of $1$-dimensional representations. 

*$D$ is isomorphic to $(\mathbb{C}^*)^r\times A$, for some finite abelian group $A$. 
I'm fine with #1-3, but I had to pause after reading #4 because I am now wondering why and how does this torsion part arise (it may be because I do not know any algebraic groups other than the ones that can (naturally) be embedded into $GL(n,\mathbb{C})$).
Example: Take $D$ to be a diagonal group with character group $\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}$. Then the coordinate ring of $D$ is $\mathbb{C}[x,x^{-1},y]/\langle y^2-1\rangle$ with $D\cong \mathbb{C}^*\times \mu_2$. 

$\mathbf{Question \;1}$: Is there a way to write $D\cong \mathbb{C}^*\times \mu_2$ in a single matrix form, i.e., embed $D$ into $GL(n,\mathbb{F})$? Or must it be separately embedded into a product of $GL_n$'s like $GL(1,\mathbb{C})\times GL(1,\mathbb{F}_2)$? 

Suppose our above $D\cong \mathbb{C}^*\times \mu_2$ acts on $\mathbb{C}^3$ by 
$$
(x,y).(a,b,c)=(xa,x^{-1}b,yc).
$$

$\mathbf{Question\; 2}$: Then do we have 
  $$
(x,y)^2.(a,b,c)=(x^2a,x^{-2}b,c) \mbox{ with } \mathbb{C}[a,b,c]^D=\mathbb{C}[ab,c^2]?
$$
$\bf{Question \; 3}$: Can you give me an example of an algebraic group which cannot be embedded into a product of $GL_n$'s? 

Thank you. 
 A: Some comments on question 2. Your answer is correct. It is a question in classical invariant theory.
$C^*\times \mu_2 \cong GL(1)\times O(1)$, act on $\mathbb{C}[\mathbb{C} \oplus \mathbb{C}^*]\otimes \mathbb{C}[\mathbb{C}]$ one each factor respectively. Classical invariant theory tell you the $GL(n)$-invariant in $\mathbb{C}[\mathbb{C}^n\oplus (\mathbb{C}^n)^*]$-the first factor is generated by $<a,b>$ (the natural paring) and the $O(n)$ invariant in $\mathbb{C}[\mathbb{C}^n]$-the second factor is generated by $c^2$.  But it is not obvious there is no further relations between the quadratic invariants. Fortunately, your case $n=1$, is in stable range, so no further relations.
A: In response to Question 3: 
The Lie group $G= \mathbb{R}\times \mathbb{R}\times S^1$ where $S^1=\{ z\in\mathbb{C}: |z|=1 \}$ with group operation 
$$
(x_1, y_1,u_1)\cdot (x_2,y_2, u_2)= (x_1 + x_2,y_1+y_2,e^{ix_1y_2}u_1 u_2)  
$$ 
is not a matrix group. 
This was obtained from Brian Hall's Lie Groups, Lie Algebras, and representations. 
