Let $G$ be a group and $N$ a normal subgroup of $G$. Must $G$ contain a subgroup isomorphic to $G/N$? My first guess is no, but by the fundamental theorem of abelian groups it is true for finite abelian groups, so finding a counterexample has been a little tough. The finite case is also interesting for me.
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3Well, $\mathbb Z$ is a finitely generated abelian group, and has no subgroup isomorphic to $\mathbb Z/2\mathbb Z$. – Lubin Aug 26 '14 at 01:10
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For a finite counterexample, take a look at the subgroup structure of the quaternions, a nonabelian group of order $8$.
By inspection, it's straightforward to verify that $Q/Z(Q)\simeq V_4$, the Klein group. However, the only subgroups of $Q$ of order $4$ are $\langle i\rangle\simeq\langle j\rangle\simeq\langle k\rangle\simeq\mathbb{Z}/(4)$. So $Q$ does not have a subgroup isomorphic to $Q/Z(Q)$.
Ben West
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thank you, am I correct that there can't be a finite abelian counterexample? – Asinomás Aug 25 '14 at 23:46
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Let $G$ be the reals $\mathbb{R}$under addition. Let $N$ be the integers $\mathbb{Z}$ under addition. $G/N$ is the group of reals modulo $1$. This has a non-zero element whose square is $1$, the equivalence class of $0.5$. So $G/N$ is not isomorphic to any subgroup of $G$.
Dan Piponi
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hmm, sorry if I'm coming off as a major noob, but doesn't the fundamental theorem of abelian groups imply any finite examples must be non-abelian? – Asinomás Aug 25 '14 at 23:41
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Haha, sorry Dan, I'm not really sure what I should do, should I find a quotient which gives me an element of order smaller than I could find in the group? – Asinomás Aug 25 '14 at 23:57