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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 "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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    $\begingroup$ Did you mean "kernel of some epimorphism $H\rightarrow G$"? $\endgroup$
    – Alex B.
    Commented Feb 27, 2015 at 12:35
  • $\begingroup$ Yes, so sorry, I corrected the question. $\endgroup$
    – grok
    Commented Feb 27, 2015 at 15:13
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    $\begingroup$ For $G$ dihedral of order $8$ there is a group $H$ of order $128$ with this property, so they can exist for nonabelian groups. $\endgroup$
    – Derek Holt
    Commented Feb 27, 2015 at 16:11
  • $\begingroup$ If $G$ has such an $H$, each quotient of $G$ has one. Do you know something about $G$ (non-abelian) simple? $\endgroup$
    – j.p.
    Commented Feb 27, 2015 at 16:33
  • $\begingroup$ @j.p., I'd love to know this for alternating groups. Thanks Derek, I'm willing to believe that $H$ always exists for nilpotent $G$. I haven't found any $H$ for $G=$ the non-nilpotent group of order 6. $\endgroup$
    – grok
    Commented Feb 27, 2015 at 18:40

1 Answer 1

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If $G$ is nonabelian simple, then $H$ cannot exist.

Lemma : Let $\cal X$ be a collection of normal subgroups of a group $H$ such that $\bigcap \cal X = 1$, and assume that $H/X$ is nonabelian simple for all $X \in \cal X$. Then $H$ is isomorphic to a direct product of nonabelian simple groups.

Proof: Let $M \triangleleft H$ be minimal such that $H/M$ is isomorphic to a direct product of nonabelian simple groups. To complete the proof, we show $M = 1$. Otherwise, there exists $X \in \cal X$ such that $M \cap X < M$, and we write $D = M \cap X$. Now $M \not\subseteq X$, so $MX > X$, and since $H/X$ is simple, it follows that $MX = G$. Then $H/D \cong M/D \times X/D$, and this is a direct product of nonabelian simple groups since $M/D \cong H/X$ is nonabelian simple and $X/D \cong H/M$ is a direct product of nonabelian simples. This contradicts the minimality of $M$.QED

Theorem: A finite group $H$ cannot be covered by a collection $\cal X$ of normal subgroups such that $H/X$ is nonabelian simple for all $X \in \cal X$.

Proof: Assume $H$ actually can be so covered, and let $D = \bigcap \cal X$. Then $D \triangleleft H$ and $H/D$ is covered by the normal subgroups $X/D$, and each of these has a nonabelian simple quotient. We can thus replace $H$ by $H/D$, so we can assume $D = 1$. The lemma then applies, and $H$ is a direct product of a collection $\cal Y$ of nonabelian simple subgroups. Then every normal subgroup of $H$ is a product of some subset of $\cal Y$, and proper normal subgroups are products of proper subsets of $\cal Y$. Now choose one nonidentity element of each member of $\cal Y$ and let $h$ be the product of the chosen elements. Then $h$ lies in no proper normal subgroup of $H$, and this is a contradiction.

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  • $\begingroup$ Thanks a lot! Followup questions: does it follow that $G$ must be solvable? Is there an $H$ for $G=S_3$ or $S_4$? $\endgroup$
    – grok
    Commented Feb 28, 2015 at 9:03
  • $\begingroup$ @grok: The follup questions you should post properly (not just in a comment). $\endgroup$
    – j.p.
    Commented Feb 28, 2015 at 14:26

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