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YES to Question 1. For an arbitrary homomorphism $\phi\colon G\to F$ of connected algebraic groups over $k$, not necessarily affine, where $k$ is a $p$-adic field or $k=\mathbb{R}$, the image $\phi(G(k))$ is closed in $F(k)$.

Proof. If $G$ is a connected $k$-group, and $X=G/H$ is a homogeneous space of $G$, then every orbit of $G(k)$ in $X(k)$ is open. Indeed, for any point $x\in X(k)$, the differential at any element $g\in G(k)$ of the map $\psi_x\colon G(k)\to X(k)\colon\ g\mapsto g\cdot x$ is surjective (we are in characteristic 0), and we can apply the Implicit Function Theorem over $k$. Since every orbit is open, we see that every orbit is closed.

Now if $\phi\colon G\to F$ is an epimorphism of connected $k$-groups, then $F$ is a homogeneous space of $G$ with respect to the action $g*f=\phi(g)f$, hence the orbit $\phi(G(k))$ of $1\in F(k)$ is open and closed in $F(k)$. If $\phi\colon G\to F$ is an arbitrary homomorphism of connected $k$-groups (not necessarily epimorphism), then the image ${\rm im}\,\phi$ is Zariski closed in $F$, see Borel's book, Cor. 1.4(a), and $G\to{\rm im}\,\phi$ is an epimorphism. We see that $\phi(G(k))$ is open and closed in the closed subgroup $({\rm im}\,\phi)(k)\subset F(k)$, hence $\phi(G(k))$ is closed in $F(k)$.

NO? to Question 2. Set $H={\rm SL}_{2,k}\times_k {\rm SL}_{2,k}$, $C=\{\pm 1 \}\times \{ \pm1\} $ $=Z/2Z\oplus Z/2Z$. Let $E$ be an elliptic curve over $k$ that has all the points of order 2 defined over $k$. We choose central embeddings $C\hookrightarrow H$ and $C\hookrightarrow E$ and set $G=(H\times_k E)/C$ with respect to the diagonal embedding. Then $G_{\rm ant}=E$, $G_{\rm aff}=H$, $S=1$, $\bar G=H$, $U=1$. In this particular case Question 2 is whether the map $$ \phi\colon H(K)\times E(K)\to G(K) $$ is surjective for some finite extension $K/k$. Since $H^1(K,H)=1,$ an easy cohomological argument reduces Question 2 in this case to the qeuestion whether for any eliptic curve the map $$ E(K)\to E(K)\colon\ x\mapsto 2x $$ can become surjective after passing to some finite extension $K/k$.

I am not an expert on elliptic curves. In a separate question you can ask to construct an elliptic curve $E$ over $k=\mathbb{Q}_p$ such that the homomorphism $$ E(K)\to E(K), \quad x\mapsto 2x $$ is not surjective for any finite extension $K/k$.

YES to Question 1. For an arbitrary homomorphism $\phi\colon G\to F$ of connected algebraic groups over $k$, not necessarily affine, where $k$ is a $p$-adic field or $k=\mathbb{R}$, the image $\phi(G(k))$ is closed in $F(k)$.

Proof. If $G$ is a connected $k$-group, and $X=G/H$ is a homogeneous space of $G$, then every orbit of $G(k)$ in $X(k)$ is open. Indeed, for any point $x\in X(k)$, the differential at any element $g\in G(k)$ of the map $\psi_x\colon G(k)\to X(k)\colon\ g\mapsto g\cdot x$ is surjective (we are in characteristic 0), and we can apply the Implicit Function Theorem over $k$. Since every orbit is open, we see that every orbit is closed.

Now if $\phi\colon G\to F$ is an epimorphism of connected $k$-groups, then $F$ is a homogeneous space of $G$ with respect to the action $g*f=\phi(g)f$, hence the orbit $\phi(G(k))$ of $1\in F(k)$ is open and closed in $F(k)$. If $\phi\colon G\to F$ is an arbitrary homomorphism of connected $k$-groups (not necessarily epimorphism), then the image ${\rm im}\,\phi$ is Zariski closed in $F$, see Borel's book, Cor. 1.4(a), and $G\to{\rm im}\,\phi$ is an epimorphism. We see that $\phi(G(k))$ is open and closed in the closed subgroup $({\rm im}\,\phi)(k)\subset F(k)$, hence $\phi(G(k))$ is closed in $F(k)$.

NO? to Question 2. Set $H={\rm SL}_{2,k}\times_k {\rm SL}_{2,k}$, $C=\{\pm 1 \}\times \{ \pm1\} $ $=Z/2Z\oplus Z/2Z$. Let $E$ be an elliptic curve over $k$ that has all the points of order 2 defined over $k$. We choose central embeddings $C\hookrightarrow H$ and $C\hookrightarrow E$ and set $G=(H\times_k E)/C$ with respect to the diagonal embedding. Then $G_{\rm ant}=E$, $G_{\rm aff}=H$, $S=1$, $\bar G=H$, $U=1$. In this particular case Question 2 is whether the map $$ \phi\colon H(K)\times E(K)\to G(K) $$ is surjective for some finite extension $K/k$. Since $H^1(K,H)=1,$ an easy cohomological argument reduces Question 2 in this case to the question whether for any elliptic curve $E$ the map $$ E(K)\to E(K)\colon\ x\mapsto 2x $$ can become surjective after passing to some finite extension $K/k$.

I am not an expert on elliptic curves. In a separate question you can ask to construct an elliptic curve $E$ over $k=\mathbb{Q}_p$ such that the homomorphism $$ E(K)\to E(K), \quad x\mapsto 2x $$ is not surjective for any finite extension $K/k$.

YES to Question 1. For an arbitrary homomorphism $\phi\colon G\to F$ of connected algebraic groups over $k$, not necessarily affine, where $k$ is a $p$-adic field or $k=\mathbb{R}$, the image $\phi(G(k))$ is closed in $F(k)$.

Proof. If $G$ is a connected $k$-group, and $X=G/H$ is a homogeneous space of $G$, then every orbit of $G(k)$ in $X(k)$ is open. Indeed, for any point $x\in X(k)$, the differential at any element $g\in G(k)$ of the map $\psi_x\colon G(k)\to X(k)\colon\ g\mapsto g\cdot x$ is surjective (we are in characteristic 0), and we can apply the Implicit Function Theorem over $k$. Since every orbit is open, we see that every orbit is closed.

Now if $\phi\colon G\to F$ is an epimorphism of connected $k$-groups, then $F$ is a homogeneous space of $G$ with respect to the action $g*f=\phi(g)f$, hence the orbit $\phi(G(k))$ of $1\in F(k)$ is open and closed in $F(k)$. If $\phi\colon G\to F$ is an arbitrary homomorphism of connected $k$-groups (not necessarily epimorphism), then $\phi(G(k))$ is open and closed in the closed subgroup $({\rm im}\,\phi)(k)\subset F(k)$, hence $\phi(G(k))$ is closed in $F(k)$.

NO to Question 2? Let $G=(SL_2\times_k A)/\{\pm1\}$, where $A$ is an elliptic curve over $k$. Then $G_{\rm ant}=A$, $G_{\rm aff}=SL_2$, $S=1$, $\bar G=SL_2$, $U=1$. In this particular case Question 2 is whether the map $$ \phi\colon SL_2(K)\times A(K)\to G(K) $$ is surjective for some finite extension $K/k$. Since $H^1(K,SL_2)=1$, we can take the quotients by $SL_2$ of the left-hand side and of the right-hand side, so in this case Question 2 is whether the map $$ A(K)\to A(K)\colon\ x\mapsto 2x $$ can be surjective for some finite extension $K/k$. I doubt that this is possible for a non-CM elliptic curve.

YES to Question 1. For an arbitrary homomorphism $\phi\colon G\to F$ of connected algebraic groups over $k$, not necessarily affine, where $k$ is a $p$-adic field or $k=\mathbb{R}$, the image $\phi(G(k))$ is closed in $F(k)$.

Proof. If $G$ is a connected $k$-group, and $X=G/H$ is a homogeneous space of $G$, then every orbit of $G(k)$ in $X(k)$ is open. Indeed, for any point $x\in X(k)$, the differential at any element $g\in G(k)$ of the map $\psi_x\colon G(k)\to X(k)\colon\ g\mapsto g\cdot x$ is surjective (we are in characteristic 0), and we can apply the Implicit Function Theorem over $k$. Since every orbit is open, we see that every orbit is closed.

Now if $\phi\colon G\to F$ is an epimorphism of connected $k$-groups, then $F$ is a homogeneous space of $G$ with respect to the action $g*f=\phi(g)f$, hence the orbit $\phi(G(k))$ of $1\in F(k)$ is open and closed in $F(k)$. If $\phi\colon G\to F$ is an arbitrary homomorphism of connected $k$-groups (not necessarily epimorphism), then the image ${\rm im}\,\phi$ is Zariski closed in $F$, see Borel's book, Cor. 1.4(a), and $G\to{\rm im}\,\phi$ is an epimorphism. We see that $\phi(G(k))$ is open and closed in the closed subgroup $({\rm im}\,\phi)(k)\subset F(k)$, hence $\phi(G(k))$ is closed in $F(k)$.

NO? to Question 2. Set $H={\rm SL}_{2,k}\times_k {\rm SL}_{2,k}$, $C=\{\pm 1 \}\times \{ \pm1\} $ $=Z/2Z\oplus Z/2Z$. Let $E$ be an elliptic curve over $k$ that has all the points of order 2 defined over $k$. We choose central embeddings $C\hookrightarrow H$ and $C\hookrightarrow E$ and set $G=(H\times_k E)/C$ with respect to the diagonal embedding. Then $G_{\rm ant}=E$, $G_{\rm aff}=H$, $S=1$, $\bar G=H$, $U=1$. In this particular case Question 2 is whether the map $$ \phi\colon H(K)\times E(K)\to G(K) $$ is surjective for some finite extension $K/k$. Since $H^1(K,H)=1,$ an easy cohomological argument reduces Question 2 in this case to the qeuestion whether for any eliptic curve the map $$ E(K)\to E(K)\colon\ x\mapsto 2x $$ can become surjective after passing to some finite extension $K/k$.

I am not an expert on elliptic curves. In a separate question you can ask to construct an elliptic curve $E$ over $k=\mathbb{Q}_p$ such that the homomorphism $$ E(K)\to E(K), \quad x\mapsto 2x $$ is not surjective for any finite extension $K/k$.

YES to Question 1. For an arbitrary homomorphism $\phi\colon G\to F$ of connected algebraic groups over $k$, not necessarily affine, where $k$ is a $p$-adic field or $k=\mathbb{R}$, the image $\phi(G(k))$ is closed in $F(k)$.

Proof. If $G$ is a connected $k$-group, and $X=G/H$ is a homogeneous space of $G$, then every orbit of $G(k)$ in $X(k)$ is open. Indeed, for any $x\in X(k)$ the differential at $1\in G(k)$ of the map $\psi_x\colon G(k)\to X(k)\colon\ g\mapsto g\cdot x$ is surjective (we are in characteristic 0), and we can apply the Implicit Function Theorem over $k$. Since every orbit is open, we see that every orbit is closed.

Now if $\phi\colon G\to F$ is an epimorphism of connected $k$-groups, then $F$ is a homogeneous space of $G$ with respect to the action $g*f=\phi(g)f$, hence the orbit $\phi(G(k))$ of $1\in F(k)$ is open and closed in $F(k)$. If $\phi\colon G\to F$ is an arbitrary homomorphism of connected $k$-groups (not necessarily epimorphism), then $\phi(G(k))$ is open and closed in the closed subgroup $({\rm im}\,\phi)(k)\subset F(k)$, hence $\phi(G(k))$ is closed in $F(k)$.

NO to Question 2? Let $G=(SL_2\times_k A)/\{\pm1\}$, where $A$ is an elliptic curve over $k$. Then $G_{\rm ant}=A$, $G_{\rm aff}=SL_2$, $S=1$, $\bar G=SL_2$, $U=1$. In this particular case Question 2 is whether the map $$ \phi\colon SL_2(K)\times A(K)\to G(K) $$ is surjective for some finite extension $K/k$. Since $H^1(K,SL_2)=1$, we can take the quotients by $SL_2$ of the left-hand side and of the right-hand side, so in this case Question 2 is whether the map $$ A(K)\to A(K)\colon\ x\mapsto 2x $$ can be surjective for some finite extension $K/k$. I doubt that this is possible for a non-CM elliptic curve.

YES to Question 1. For an arbitrary homomorphism $\phi\colon G\to F$ of connected algebraic groups over $k$, not necessarily affine, where $k$ is a $p$-adic field or $k=\mathbb{R}$, the image $\phi(G(k))$ is closed in $F(k)$.

Proof. If $G$ is a connected $k$-group, and $X=G/H$ is a homogeneous space of $G$, then every orbit of $G(k)$ in $X(k)$ is open. Indeed, for any point $x\in X(k)$, the differential at any element $g\in G(k)$ of the map $\psi_x\colon G(k)\to X(k)\colon\ g\mapsto g\cdot x$ is surjective (we are in characteristic 0), and we can apply the Implicit Function Theorem over $k$. Since every orbit is open, we see that every orbit is closed.

Now if $\phi\colon G\to F$ is an epimorphism of connected $k$-groups, then $F$ is a homogeneous space of $G$ with respect to the action $g*f=\phi(g)f$, hence the orbit $\phi(G(k))$ of $1\in F(k)$ is open and closed in $F(k)$. If $\phi\colon G\to F$ is an arbitrary homomorphism of connected $k$-groups (not necessarily epimorphism), then $\phi(G(k))$ is open and closed in the closed subgroup $({\rm im}\,\phi)(k)\subset F(k)$, hence $\phi(G(k))$ is closed in $F(k)$.

NO to Question 2? Let $G=(SL_2\times_k A)/\{\pm1\}$, where $A$ is an elliptic curve over $k$. Then $G_{\rm ant}=A$, $G_{\rm aff}=SL_2$, $S=1$, $\bar G=SL_2$, $U=1$. In this particular case Question 2 is whether the map $$ \phi\colon SL_2(K)\times A(K)\to G(K) $$ is surjective for some finite extension $K/k$. Since $H^1(K,SL_2)=1$, we can take the quotients by $SL_2$ of the left-hand side and of the right-hand side, so in this case Question 2 is whether the map $$ A(K)\to A(K)\colon\ x\mapsto 2x $$ can be surjective for some finite extension $K/k$. I doubt that this is possible for a non-CM elliptic curve.