Assuming $G$ finite, $A$ is of the form $k[H]$ for some subgroup of $G$ if and only if $A$ is a sub-bialgebra (and since $G$ is finite, if and only if is Hopf subalgebra).

Clearly $k[H]$ is a Hopf subalgebra of $k[G]$, but if you take any subalgebra $A$, the fact that $A$ is also a subcoalgebra means that $A$ is a subcomodule of $k[G]$, hence $G$-graded, so, the $G$-homogeneous components are $1$-dimentionals and $A$ is generated by group-like elements. But a subcoalgebra of the form $k[X]$ with $X\subseteq G$ is subalgebra only when $X$ is a subgroup.

(By the way, if $G$ is infinite everything works fine except that $H$ is maybe a submonoid and not a subgroup).

An alternative proof avoiding the grading/comodule argument is the following: $A$ being subcoalgebra means $A^*$ is an algebra quotient of $k[G]^*=k^G$= the algebra of functions from G to k. But it is clear that any quotient of $k^G$ identifies with $k^X$ with $X\subseteq G$. The rest of the argument is the same.

Yes, there is a criterium. Assuming $G$ finite, $A$ is of the form $k[H]$ for some subgroup of $G$ if and only if $A$ is a sub-bialgebra (and since $G$ is finite, if and only if is Hopf subalgebra).

Clearly $k[H]$ is a Hopf subalgebra of $k[G]$, but if you take any subalgebra $A$, the fact that $A$ is also a subcoalgebra means that $A$ is a subcomodule of $k[G]$, hence $G$-graded, so, the $G$-homogeneous components are $1$-dimentionals and $A$ is generated by group-like elements. But a subcoalgebra of the form $k[X]$ with $X\subseteq G$ is subalgebra only when $X$ is a subgroup.

(By the way, if $G$ is infinite everything works fine except that $H$ is maybe a submonoid and not a subgroup).

An alternative proof avoiding the grading/comodule argument is the following: $A$ being subcoalgebra means $A^*$ is an algebra quotient of $k[G]^*=k^G$= the algebra of functions from G to k. But it is clear that any quotient of $k^G$ identifies with $k^X$ with $X\subseteq G$. The rest of the argument is the same.

Assuming G finite, A is of the form k[H] for some subgroup of G if and only if A is a sub-bialgebra (and since G is finite, if and only if is Hopf subalgebra).

Clearly k[H] is a Hopf subalgebra of k[G], but if you take any subalgebra A, the fact that A is also a subcoalgebra means that A is a subcomodule of k[G], hence G-graded, so, the G-homogeneous components are 1-dimentionals and A is generated by group-like elements. But a subcoalgebra of the form k[X] with $X\subseteq G$ is subalgebra only when X is a subgroup.

(By the way, if G is infinite everything works fine except that H is maybe a submonoid and not a subgroup).

An alternative proof avoiding the grading/comodule argument is the following: A being subcoalgebra means $A^*$ is an algebra quotient of $k[G]^*=k^G$= the algebra of functions from G to k. But it is clear that any quotient of $k^G$ identifies with $k^X$ with $X\subseteq G$. The rest of the argument is the same.

Assuming $G$ finite, $A$ is of the form $k[H]$ for some subgroup of $G$ if and only if $A$ is a sub-bialgebra (and since $G$ is finite, if and only if is Hopf subalgebra).

Clearly $k[H]$ is a Hopf subalgebra of $k[G]$, but if you take any subalgebra $A$, the fact that $A$ is also a subcoalgebra means that $A$ is a subcomodule of $k[G]$, hence $G$-graded, so, the $G$-homogeneous components are $1$-dimentionals and $A$ is generated by group-like elements. But a subcoalgebra of the form $k[X]$ with $X\subseteq G$ is subalgebra only when $X$ is a subgroup.

(By the way, if $G$ is infinite everything works fine except that $H$ is maybe a submonoid and not a subgroup).

An alternative proof avoiding the grading/comodule argument is the following: $A$ being subcoalgebra means $A^*$ is an algebra quotient of $k[G]^*=k^G$= the algebra of functions from G to k. But it is clear that any quotient of $k^G$ identifies with $k^X$ with $X\subseteq G$. The rest of the argument is the same.