A regular embedding of a connected reductive linear algebraic group $G$ defined over $\mathbb{F}_q$ is a morphism $\varphi : G \rightarrow G'$ of algebraic groups which is a closed immersion where $G'$ is a connected reductive linear algebraic group defined over $\mathbb{F}_q$ with connected centre and $\varphi(G)$ contains the derived subgroup of $G'$.

For example, if $T$ is a torus and $\iota: Z(G) \rightarrow T$ an isomorphism onto its image, then mapping $G$ into the quotient of $G \times T$ by $Z = \{(z,\iota(z)^{-1})\:|\: z \in Z(G)\}$ gives a regular embedding.

I want to prove the following statement:

Given two regular embeddings, say $G \rightarrow G'$ and $G \rightarrow G''$, there exist regular embeddings $G' \rightarrow G'''$ and $G'' \rightarrow G'''$ making the resulting square of morphisms commutative.

I believe I can prove this in the special case where $G'$ and $G''$ are of the special form $(G \times T)/Z$ like above. For if $G'$ is such a quotient of $G \times T'$ and $G''$ is such a quotient of $G \times T''$, then I can take a similar quotient of $G \times T' \times T''$ for $G'''$ and this seems to work as far as I can tell.

However, I fail to understand the general case completely. Note that if $\varphi : G \rightarrow G'$ is a regular embedding, then $G' = Z(G')\varphi(G)$ and $Z(G')$ is a torus. We then have a bijective morphism $(G \times Z(G'))/Z \rightarrow G'$ where $Z$ is like above using $\varphi|_{Z(G)} : Z(G) \rightarrow Z(G')$. But this bijection is not necessarily an isomorphism (at least I could not prove that).

For reference, this is Excercise 2 in Chapter 15 of "Representation theory of finite Reductive groups" by Cabanes and Enguehard. I found the statement also in several other papers but without any hint towards a proof. Any help is appreciated.

A regular embedding of a connected reductive linear algebraic group $G$ defined over $\mathbb{F}_q$ is a morphism $\varphi : G \rightarrow G'$ of algebraic groups which is a closed immersion where $G'$ is a connected reductive linear algebraic group defined over $\mathbb{F}_q$ with connected centre and $\varphi(G)$ contains the derived subgroup of $G'$.

For example, if $T$ is a torus and $\iota: Z(G) \rightarrow T$ an isomorphism onto its image, then mapping $G$ into the quotient of $G \times T$ by $Z = \{(z,\iota(z)^{-1})\:|\: z \in Z(G)\}$ gives a regular embedding.

I want to prove the following statement:

Given two regular embeddings, say $G \rightarrow G'$ and $G \rightarrow G''$, there exist regular embeddings $G' \rightarrow G'''$ and $G'' \rightarrow G'''$ making the resulting square of morphisms commutative.

I believe I can prove this in the special case where $G'$ and $G''$ are of the special form $(G \times T)/Z$ like above. For if $G'$ is such a quotient of $G \times T'$ and $G''$ is such a quotient of $G \times T''$, then I can take a similar quotient of $G \times T' \times T''$ for $G'''$ and this seems to work as far as I can tell.

However, I fail to understand the general case completely. Note that if $\varphi : G \rightarrow G'$ is a regular embedding, then $G' = Z(G')\varphi(G)$ and $Z(G')$ is a torus. We then have a bijective morphism $(G \times Z(G'))/Z \rightarrow G'$ where $Z$ is like above using $\varphi|_{Z(G)} : Z(G) \rightarrow Z(G')$. But this bijection is not necessarily an isomorphism (at least I could not prove that).

For reference, this is Exercise 2 in Chapter 15 of "Representation Theory of Finite Reductive Groups" by Cabanes and Enguehard (Cambridge, 2004). I found the statement also in several other papers but without any hint towards a proof. Any help is appreciated.