Let $G$ be a reductive group over the finite field $\mathbb{F}\_q$. Then a *regular* embedding of $G$ is an $\mathbb{F}\_q$-rational embedding $\iota \colon G \rightarrow G'$ into a second reductive group over $\mathbb{F}\_q$, such that $\iota$ maps the derived group of $G$ isomorphically onto that of $G'$ and the center of $G'$ is connected. Moreover, a regular embedding is called *smooth regular*, if additionally the center of $G'$ is smooth. It is well-known that smooth regular embeddings always exist (see for example [1,Theorem 4.5],[2, Lemma 6.5]).


**Question.** Is it true that for any $G$ a smooth regular embedding $\iota \colon G \rightarrow G'$ exists, with the additional property that the center of $G'$ is an induced torus?


It seems quite plausible, but I could not find any reference (which is a bit annoying).


Note that for the somewhat dual notion of a *z-extension* $\widetilde G \rightarrow G$ it can be arranged -- and is even part of the definition of a z-extension (cf. [3,§1]) -- that ${\rm ker}(\widetilde{G} \rightarrow G)$ is induced.


References:


[1] Martin, B.: Étale slices for representation varieties in characteristic p.


[2] Taylor, J.: The Structure of Root Data and Smooth Regular Embeddings of Reductive Groups.


[3] Kottwitz, R.: Rational conjugacy classes in reductive groups.



Regular embeddings of a reductive groups with induced center
The answer is **Yes**.


Let $G\hookrightarrow G'$ be a smooth regular embedding. We write $Z(G')$ for the center of $G'$, which is an $F$-torus, where $F={\Bbb F}\_q$.
We construct a regular embedding $G'\hookrightarrow G''$ such that $Z'':=Z(G'')$ is an induced torus. We write $S=(G,G)=(G',G')$ for the derived group of $G$ and $G'$.


Consider the natural surjective homomorphism
$$\varphi\colon Z'\times\_F S\to G',\quad (z,s)\mapsto z^{-1}\cdot s.$$
Set $K=\ker\varphi=\big\{(z^{-1},z)\mid z\in Z'\cap Z(S)\big\}$. Then we have a canonical isomorphism
$$(Z'\times\_F S)/K\to G'.$$
Since $Z'=Z(G')$, we see that $Z'\supset Z(S)$, whence
$K=\big\{(z^{-1},z)\mid z\in Z(S)\big\}$.


Write $X'$ for the character group of $Z'$. Then $X'$ is a ${\rm Gal}(\overline F/F)$-module.
Clearly, there exists a surjective homomorphism $\alpha\colon X''\twoheadrightarrow X'$, where $X''$ is a permutation ${\rm Gal}(\overline F/F)$-module. Let $Z''$ denote the $F$-torus with character group $X''$, which is an induced torus. Then we have an injective homomorphism $\alpha^\*\colon\, Z'\hookrightarrow Z''$.


Consider the induced injective homomorphism
$$\alpha^{\*\*}\colon\, Z'\times\_F S\hookrightarrow Z''\times\_F S$$
and the induced injective homomorphism
$$\alpha^\vee\colon\, G'= (Z'\times\_F S)/K\hookrightarrow
(Z''\times\_F S)/\alpha^{\*\*}(K)=:G''.$$
Then $$\alpha^{\*\*}(K)=\big\{(\alpha^\*(z)^{-1},z)\mid z\in Z(S)\big\}.$$
It follows that the homomorphism $Z''\to Z(G'')$
is an isomorphism and that $\alpha^\vee$ is a regular embedding.
The composite injective homomorphism
$$ G\hookrightarrow G'\hookrightarrow G''$$
is a desired smooth regular embedding for which $Z(G'')$ is an induced torus.


