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It suffices to show that $u$ lies in the unipotent radical of a "canonical" parabolic subgroup $P$ (i.e., one stable under all automorphisms of $G$ that preserve $u$, or even just preserve the cyclic subgroup generated by $u$). Indeed, once this is proved, such a $P$ is stable under $U$-conjugation on $G$ (as that leaves the central $u$ in $U(k)$ invariant), so by the self-normalizing property of parabolic subgroups it would follow that $U \subset P$. But $P \ne G$ (since $R_u(P)$ contains $u \ne 1$ whereas we arranged that $G$ is reductive), so we can use dimension induction to conclude.

We are left with a serious problem concerning a single unipotent element $u \in G(k)$ (with $G$ a connected linear algebraic group): find a "canonical" parabolic subgroup $P$ of $G$ that contains $u$. (This may not seem like progress, as we have reduced the initial claim to a problem involving a cyclic group, but we will soon see that this is in fact genuine progress.) The weaker claim that $u$ lies in the unipotent radical of some parabolic subgroup, or equivalently in the unipotent radical of some Borel subgroup, is immediate from Grothendieck's important theorem that $G(k)$ is covered by the subgroups $B(k)$ for Borel subgroups $B$ of $G$ (a result that is proved near the end of section 14 of Borel's textbook but possibly not stated as an official result there; it can be extracted nonetheless). We are going to actually use Grothendieck's theorem to get things started.

By Grothendieck's theorem, $u$ lies in $R_u(B)$, so the cyclic subgroup generated by $u$ is "embeddable" (in the terminology of Borel--Tits), meaning that it does occur inside some connected unipotent subgroup (equivalently, in the unipotent radical of a Borel subgroup). The problem is to show that there is a canonical parabolic subgroup whose unipotent radical contains that cyclic group (or equivalently, contains its unipotent Zariski closure). Rather than focus on cyclic groups, consider more generally a closed (perhaps disconnected) unipotent subgroup $U$ of $G$ that is "embeddable"; i.e., is contained in a connected unipotent subgroup. It suffices to prove the following refinement of the previous theorem:

We will show that $u$ lies in the unipotent radical of a parabolic subgroup $P$ that is "canonical" in terms of $u$ (i.e., one stable under all automorphisms of $G$ that preserve $u$, or even just preserve the cyclic subgroup generated by $u$). Once this is proved, such a $P$ is stable under $U$-conjugation on $G$ (as that leaves the central $u$ in $U(k)$ invariant), so by the self-normalizing property of parabolic subgroups it would follow that $U \subset P$. But $P \ne G$ (since $R_u(P)$ contains $u \ne 1$ whereas we arranged that $G$ is reductive), so we can use dimension induction to conclude that $U$ lies in a connected unipotent subgroup of $P$, but without canonicity in $(G,U)$. However, we gain the property that $U$ is "embeddable" (in the terminology of Borel--Tits), meaning that it does occur inside some connected unipotent subgroup (equivalently, in the unipotent radical of a Borel subgroup).

Observe that $u$ lies in the unipotent radical of some parabolic subgroup, or equivalently in the unipotent radical of some Borel subgroup: this is immediate from Grothendieck's important theorem that $G(k)$ is covered by the subgroups $B(k)$ for Borel subgroups $B$ of $G$ (a result that is proved near the end of section 14 of Borel's textbook but possibly not stated as an official result there; it can be extracted nonetheless). Thus, the Zariski closure of the cyclic subgroup generated by $u$ is "embeddable". Moreover, we saw above via the dimension induction that once this cyclic subgroup is realized inside the unipotent radical of a "canonically associated" parabolic subgroup of $G$, then $U$ itself is "embeddable". Hence, by two applications of the following result (which is to be regarded as a finer version of Grothendieck's theorem, but whose applicability in our inductive strategy ultimately rests on Grothendieck's theorem) we would be done:

This is a (rather) expanded version of Yves' answer, because for the specific focus of the question one can understand the proof without getting into many other technicalities of the Borel--Tits paper (which has bigger fish to fry, related to fields of definition and much more). We will explain the proof of the following result:

${\bf Theorem}$ (Borel--Tits): In a connected linear algebraic group $G$ over an algebraically closed field $k$, every subgroup $H \subset G(k)$ consisting of unipotent elements is contained in a "canonical" connected unipotent subgroup (i.e., one that is stable under all automorphisms of $G$ that carries $H$ onto itself).

Proof: Since the Zariski closure of $H$ is unipotent (but perhaps disconnected), we can replace $H$ with its Zariski closure and instead work with unipotent closed subgroups $U$ in $G$ (e.g., finite ones when in positive characteristic). We induct on the dimension of $G$ (dimension 0 being trivial). We can also replace $U$ with $U \cdot R_u(G)$ so that $U$ contains $R_u(G)$, and if $R_u(G) \ne 1$ then we can pass to $G/R_u(G)$ to drop the dimension and conclude, so we can assume that $G$ is reductive. We can assume $U \ne 1$, so by nilpotence of unipotent algebraic groups there is a nontrivial central element $u$ in $U(k)$.

We are left with a serious problem concerning a single unipotent element $u \in G(k)$ (with $G$ a connected linear algebraic group): find a "canonical" parabolic subgroup $P$ of $G$ that contains $u$. (This may not seem like progress, as we have reduced the initial grand claim with canonical parabolics to a problem involving a cyclic group, but we will soon see that this is in fact genuine progress.) The weaker claim that $u$ lies in the unipotent radical of some parabolic subgroup, or equivalently in the unipotent radical of some Borel subgroup, is immediate from Grothendieck's important theorem that $G(k)$ is covered by the subgroups $B(k)$ for Borel subgroups $B$ of $G$ (a result that is proved near the end of section 14 of Borel's textbook but possibly not stated as an official result there; it can be extracted nonetheless). We are going to actually use Grothendieck's theorem to get things started.

${\bf Refined}$ ${\bf Theorem}$ (Borel--Tits). Any embeddable closed unipotent subgroup $U$ of a connected linear algebraic group $G$ is contained in the unipotent radical of a "canonically associated" parabolic subgroup $P$. (As before, "canonical" means that $P$ is stable under all automorphisms of $G$ that preserve $U$.)

[Note that this result is interesting even when $U$ is connected, and in particular it is interesting in characteristic 0. Also, the embeddable hypothesis is automatic once this is all over! The proof gives some "canonical" $P$, but it is not uniquely determined by the stated conditions in general, as we see by considering $G = \prod G_i$ for several pairwise non-isomorphic simple semi simple $G_i$ with $U \subset G_1$. I don't know offhand whether one can do better when $G$ is simple.]

Proof: To construct $P$, we will work with the normalizer $N(U)$, which might be non-unipotent and even more disconnected. Consider the unipotent radical of $N(U)$ (i.e., maximal unipotent connected normal subgroup) anyway. Note that $R_u(N(U))$ trivially contains $U^0$, but it isn't clear if it contains $U$ or not. So following Borel--Tits (see 2.4 in their paper), let's put $U$ back in: define the closed subgroup $$L(U) = U \cdot R_u(N(U))$$ that is unipotent and does contain $U$. By the "canonicity" of $L(U)$ in terms of $U$, any automorphism of $G$ that carries $U$ back to itself must do the same for $L(U)$. Also, just as we assumed that $U$ is embeddable (i.e., lies in the unipotent radical of a Borel), we claim the same holds for $L(U)$. This amounts to checking that $L(U)$ has a fixed point on the projective variety of Borel subgroups of $G$. By hypothesis $U$ has a fixed point, so the closed (!) locus of Borels containing $U$ is a non-empty projective variety (perhaps disconnected). Certainly $N(U)$ acts on that locus, so the connected solvable $R_u(N(U))$ does too. By the Borel fixed point theorem, $R_u(N(U))$ has a fixed point in that locus, so we have found a Borel subgroup of $G$ that contains $L(U)$. Hence, we lose nothing by replacing $U$ with $L(U)$.

This is a (rather) expanded version of Yves' answer, because for the specific focus of the question one can understand the proof without getting into many other technicalities of the Borel--Tits paper (which has bigger fish to fry, related to fields of definition and much more). We will explain the proof of the following result (which is given over perfect fields rather than just algebraically closed fields so that we include finite fields):

${\bf Theorem}$ (Borel--Tits): In a connected linear algebraic group $G$ over a perfect field $k$, every subgroup $H \subset G(k)$ consisting of unipotent elements is contained in a "canonical" connected unipotent smooth closed $k$-subgroup (i.e., one whose formation is compatible with respect to $k$-isomorphisms in the pair $(G,H)$ and perfect extension on $k$).

Proof: Due to the canonicity claim applied in the setting of Galois descent relative to $\overline{k}$ over $k$, we may and do assume $k$ is algebraically closed (and the compatibility with respect to any further perfect extension of the ground field will be clear from the construction). Since the Zariski closure of $H$ is (smooth and) unipotent (but perhaps disconnected), we can replace $H$ with its Zariski closure and instead work with unipotent smooth closed subgroups $U$ in $G$ (e.g., finite constant ones when in positive characteristic). We induct on the dimension of $G$ (dimension 0 being trivial). We can also replace $U$ with $U \cdot R_u(G)$ so that $U$ contains $R_u(G)$, and if $R_u(G) \ne 1$ then we can pass to $G/R_u(G)$ to drop the dimension and conclude, so we can assume that $G$ is reductive. We can assume $U \ne 1$, so by nilpotence of unipotent algebraic groups there is a nontrivial central element $u$ in $U(k)$.

We are left with a serious problem concerning a single unipotent element $u \in G(k)$ (with $G$ a connected linear algebraic group): find a "canonical" parabolic subgroup $P$ of $G$ that contains $u$. (This may not seem like progress, as we have reduced the initial claim to a problem involving a cyclic group, but we will soon see that this is in fact genuine progress.) The weaker claim that $u$ lies in the unipotent radical of some parabolic subgroup, or equivalently in the unipotent radical of some Borel subgroup, is immediate from Grothendieck's important theorem that $G(k)$ is covered by the subgroups $B(k)$ for Borel subgroups $B$ of $G$ (a result that is proved near the end of section 14 of Borel's textbook but possibly not stated as an official result there; it can be extracted nonetheless). We are going to actually use Grothendieck's theorem to get things started.

${\bf Refined}$ ${\bf Theorem}$ (Borel--Tits). Any embeddable unipotent smooth closed subgroup $U$ of a connected linear algebraic group $G$ over a perfect field $k$ is contained in the unipotent radical of a "canonically associated" parabolic $k$-subgroup $P$. (As before, "canonical" means that the formation of $P$ is compatible with respect to $k$-isomorphisms in the pair $(G,U)$ and perfect extension on $k$.)

[Note that this result is interesting even when $U$ is connected, and it is interesting in characteristic 0. Also, the embeddable hypothesis is automatic once this is all over, so we see that the unipotent radicals of the minimal parabolic $k$-subgroups are precisely the maximal unipotent smooth closed $k$-subgroups; in particular, if $G$ has no proper parabolic $k$-subgroups then $G(k)$ contains no nontrivial unipotent smooth closed $k$-subgroups. The proof gives a "canonical" $P$, but it is not uniquely determined by the stated conditions in general, as we see by considering $G = \prod G_i$ for several pairwise non-isomorphic $k$-simple semisimple $G_i$ with $U \subset G_1$. I don't know offhand whether one can do better when $G$ is $k$-simple or absolutely simple.]

Proof: By Galois descent from $\overline{k}$ we may assume $k$ is algebraically closed (and it will be clear from the construction that we have compatibility with any further perfect extension of the ground field). To construct $P$, we will work with the normalizer $N(U)$, which might be non-unipotent and even more disconnected. Consider the unipotent radical of $N(U)$ (i.e., maximal unipotent connected normal subgroup) anyway. Note that $R_u(N(U))$ trivially contains $U^0$, but it isn't clear if it contains $U$ or not. So following Borel--Tits (see 2.4 in their paper), let's put $U$ back in: define the closed subgroup $$L(U) = U \cdot R_u(N(U))$$ that is unipotent and does contain $U$. By the "canonicity" of $L(U)$ in terms of $U$, any automorphism of $G$ that carries $U$ back to itself must do the same for $L(U)$. Also, just as we assumed that $U$ is embeddable (i.e., lies in the unipotent radical of a Borel), we claim the same holds for $L(U)$. This amounts to checking that $L(U)$ has a fixed point on the projective variety of Borel subgroups of $G$. By hypothesis $U$ has a fixed point, so the closed (!) locus of Borels containing $U$ is a non-empty projective variety (perhaps disconnected). Certainly $N(U)$ acts on that locus, so the connected solvable $R_u(N(U))$ does too. By the Borel fixed point theorem, $R_u(N(U))$ has a fixed point in that locus, so we have found a Borel subgroup of $G$ that contains $L(U)$. Hence, we lose nothing by replacing $U$ with $L(U)$.

[Note that this result is interesting even when $U$ is connected, and in particular it is interesting in characteristic 0. Also, the embeddable hypothesis is automatic once this is all over! Thus, for example, since $N(U)$ contains that $P$, we see that $N(U)$ is always parabolic with $U \subset R_u(N(U))$ (so in particular $N(U)$ is connected). Thus, $N(U)$ itself is the unique maximal choice among parabolics that satisfy the properties for $P$, after the proof is over. The proof gives some other $P$, generally not $N(U)$, as we see by considering $G = \prod G_i$ for several pairwise non-isomorphic simple semi simple $G_i$ with $U \subset G_1$.]

[Note that this result is interesting even when $U$ is connected, and in particular it is interesting in characteristic 0. Also, the embeddable hypothesis is automatic once this is all over! The proof gives some "canonical" $P$, but it is not uniquely determined by the stated conditions in general, as we see by considering $G = \prod G_i$ for several pairwise non-isomorphic simple semi simple $G_i$ with $U \subset G_1$. I don't know offhand whether one can do better when $G$ is simple.]