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A common mistake for beginning group theory students is the belief that a quotient of a group $G$ is necessarily isomorphic to a subgroup of $G$. Is there a characterization of the groups in which this property holds?

If this question is too broad, I might ask if such a characterization exists for $p$-groups.


History: I originally posed the opposite question, regarding groups for which $\exists N\unlhd G\,:\, \not\exists H \unlhd G\, \text{ s.t. } H \cong G/N$, and crossposted this to MO. I received an answer there to the (now omitted) peripheral question about probability, which shows that most finite groups probably have this property. After this, I changed the question to its current state, as this smaller collection of groups is more likely to be characterizable.

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    how do you define the probability space?2012-12-30
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    Note that all quasisimple groups have the property.2012-12-30
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    All finite abelian groups *don't* have the property, and non-abelian groups of order $\,p^3\,\,,\,\,p$ a prime have it.2013-01-15
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    @DonAntonio: Are you sure about your statement for non-abelian groups of order $p^3$? $Q_8$ has the property, but a non-abelian group of order $p^3$ and exponent $p$ (implying $p>2$) doesn't have it.2013-01-16
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    I made a mess with "the property", which by the OP must be that there exists a subgroup such that non-homomorphic image of the grous is isomorphic to it, so you are right: $\,Q_8\,$ does have it taking $\,\langle i\rangle\cong C_4\leq Q_8\,$. I confused things this since "the property" in this case is defined by negation: NOT having a subgroup...etc. Thanks.2013-01-16
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    There is a discussion of this here: http://groupprops.subwiki.org/wiki/Quotient_group_need_not_be_isomorphic_to_any_subgroup2013-03-11
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    A rather trivial restatement of the property, but with obvious routes to generalization: consider all subnormal series $1 =H_0 \triangleleft H_1 \triangleleft \cdots \triangleleft H_n = G$ of $G$. The property now says that the set of top factors $G / H_{n-1}$ has an element that is not isomorphic to any in the set of bottom factors $H_1 / 1$. Are there any results on when composition factors are freely permutable?2013-03-13
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    @Alexander: However, a genuinely quasisimple group $G$ , that is one with a non-trivial center, has a homomorphic image which is not a subgroup, namely $G/Z(G).$2013-04-13
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    If $G$ is a free group of rank $\ge 2$ then $G'$ has the required property.2013-05-18
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    @yatima2975 I don't think that is in fact a restatement unless every subgroup $H$ can be put into a subnormal series. This won't be the case in general, for instance in any finite non-nilpotent group.2013-05-22
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    Every torsion-free hyperbolic group $G$ has such a subgroup $N$ *if* all hyperbolic groups are residually finite. This is because all hyperbolic groups being residually finite is equivalent to all hyperbolic groups having a subgroup of finite index. If $G$ is torsion-free hyperbolic and $N$ is a finite-index subgroup then $G/N$ is torsion and so cannot embed into $G$.2013-07-02
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    Take any vector space $V$, and $W \subset V$ a subspace of $V$, and let $W'$ be the orthogonal complement of $W$, $W' \cong V/W$.2018-11-27

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