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Let $G$ be a finite group. When does there exist a finite group $H$ such that every $h\in H$ is in the kernel of some epimorphism $H\to G$?

This is well-known to be true for $G$ abelian, for example $H=G\times G$ works. I would like very much to know such an $H$ for non-abelian $G$, e.g. for symmetric groups $G$.

Note that this is related to question "http"https://mathoverflow.net/questions/185508/normal-covering-of-a-finite-group"; there, it is discussed when $H$ is a union of proper normal subgroups (but not necessarily all with quotient $G$).

Let $G$ be a finite group. When does there exist a finite group $H$ such that every $h\in H$ is in the kernel of some epimorphism $H\to G$?

This is well-known to be true for $G$ abelian, for example $H=G\times G$ works. I would like very much to know such an $H$ for non-abelian $G$, e.g. for symmetric groups $G$.

Note that this is related to question "http://mathoverflow.net/questions/185508/normal-covering-of-a-finite-group"; there, it is discussed when $H$ is a union of proper normal subgroups (but not necessarily all with quotient $G$).

Let $G$ be a finite group. When does there exist a finite group $H$ such that every $h\in H$ is in the kernel of some epimorphism $H\to G$?

This is well-known to be true for $G$ abelian, for example $H=G\times G$ works. I would like very much to know such an $H$ for non-abelian $G$, e.g. for symmetric groups $G$.

Note that this is related to question "https://mathoverflow.net/questions/185508/normal-covering-of-a-finite-group"; there, it is discussed when $H$ is a union of proper normal subgroups (but not necessarily all with quotient $G$).

Typo in the definition corrected, thanks!
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grok
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Let $G$ be a finite group. When does there exist a finite group $H$ such that every $h\in H$ is in the kernel of some epimorphism $G\to H$$H\to G$?

This is well-known to be true for $G$ abelian, for example $H=G\times G$ works. I would like very much to know such asan $H$ for non-abelian $G$, e.g. for symmetric groups $G$.

Note that this is related to question "http://mathoverflow.net/questions/185508/normal-covering-of-a-finite-group"; there, it is discussed when $H$ is a union of proper normal subgroups (but not necessarily all with quotient $G$).

Let $G$ be a finite group. When does there exist a finite group $H$ such that every $h\in H$ is in the kernel of some epimorphism $G\to H$?

This is well-known to be true for $G$ abelian, for example $H=G\times G$ works. I would like very much to know such as $H$ for non-abelian $G$, e.g. for symmetric groups $G$.

Note that this is related to question "http://mathoverflow.net/questions/185508/normal-covering-of-a-finite-group"; there, it is discussed when $H$ is a union of proper normal subgroups (but not necessarily all with quotient $G$).

Let $G$ be a finite group. When does there exist a finite group $H$ such that every $h\in H$ is in the kernel of some epimorphism $H\to G$?

This is well-known to be true for $G$ abelian, for example $H=G\times G$ works. I would like very much to know such an $H$ for non-abelian $G$, e.g. for symmetric groups $G$.

Note that this is related to question "http://mathoverflow.net/questions/185508/normal-covering-of-a-finite-group"; there, it is discussed when $H$ is a union of proper normal subgroups (but not necessarily all with quotient $G$).

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grok
grok
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Covering finite groups by kernels

Let $G$ be a finite group. When does there exist a finite group $H$ such that every $h\in H$ is in the kernel of some epimorphism $G\to H$?

This is well-known to be true for $G$ abelian, for example $H=G\times G$ works. I would like very much to know such as $H$ for non-abelian $G$, e.g. for symmetric groups $G$.

Note that this is related to question "http://mathoverflow.net/questions/185508/normal-covering-of-a-finite-group"; there, it is discussed when $H$ is a union of proper normal subgroups (but not necessarily all with quotient $G$).