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If $H <G $ are groups and H is abelian, do we get an injection from H into $G/[G,G] $?

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  • $\begingroup$ Take $G$ perfect (e.g., (nonabelian) simple) and $H$ cyclic. $\endgroup$ – anomaly May 20 '16 at 17:07
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    $\begingroup$ Abstractly, this fails to happen because abelianization is a left adjoint but not a right adjoint. Left adjoints preserve epimorphisms but not necessarily monomorphisms in general. $\endgroup$ – Qiaochu Yuan May 20 '16 at 17:20
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No. Take $G=S_3$. The abelianization of $G$ is $C_2$. But $C_3$ is a subgroup of $S_3$ and certainly does not inject into $C_2$.

The correct universal property of the abelianization is as follows:

Let $G$ be a group, let $H$ be an abelian group and let $\phi\colon G\to H$ be a homomorphism. Then there is a unique homomorphism $\hat{\phi}\colon G/[G,G]\to H$ such that $\phi$ factors as: $$ \phi= G\xrightarrow{\pi}G/[G,G]\xrightarrow{\hat\phi} H $$

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In fact, this happens all the time for a nonabelian group $G$. If $G$ is not abelian, then its commutator subgroup $[G,G]$ is nontrivial. Take any $x\in[G,G]$ different from $1$, and let $H$ be the subgroup generated by $x$. Then $H$ is abelian but maps to $0$ in the quotient $G/[G,G]$!

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