fibred product of groups of multiplicative type (or, more generally, of linear algebraic groups) Let $M, M', M''$ be $k$-groups of multiplicative type, and let $M' \to M$ and $M'' \to M$ be morphisms of group schemes. 

Is the fibred product $M' \times_M M''$ a $k$-group of multiplicative type?

More generally,

If $G, G', G''$ are $k$-linear alg groups with maps $G' \to G$ and $G'' \to G$, is $G' \times_G G''$ a closed subgroup of $G' \times_k G''$?

 A: For convenience, I will call the maps $\varphi' : G' \to G$ and $\varphi'': G'' \to G$. The fiber product is defined by the property that for any scheme $T$,
$$
(G' \times_G G'')(T) \cong G'(T) \times_{G(T)} G''(T)
$$
where the isomorphism is natural and $G(T)$ denotes the set of morphisms $T \to G$. Now, let us look at the right hand side. As sets, the fiber product is just the subset of the product consisting of pairs $(g',g'')$ such that 
$$
\varphi'(T)(g') = \varphi''(T)(g'')
$$
Now, $G'(T), G(T)$, and $G''(T)$ are all groups since $G',G$ and $G''$ are algebraic groups and $\varphi'(T)$ and $\varphi''(T)$ are group homomorphisms. Therefore the subset 
$$
G'(T) \times_{G(T)} G''(T) = \ker(\varphi(T): G'(T) \times G''(T) \to G(T))
$$ 
where $\varphi : G' \times G'' \to G$ is defined by $\varphi(g',g'') = \varphi'(g')\varphi''(g'')^{-1}$. Clearly, this equality is natural. Therefore, by Yoneda's lemma, there is a unique isomorphism
$$
G' \times_G G'' \cong \ker(\varphi: G' \times G'' \to G).
$$
Since kernels of homomorphisms between algebraic groups are closed subgroups, then $G' \times_G G''$ is indeed a closed subgroup of $G' \times G''$. 
