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Let $\phi,\psi:G\to H$ be a pair of group homomorphisms. Their equalizer is the subgroup $C\le G$ given by $$ C=\{ g\in G \mid \phi(g)=\psi(g)\} $$ My question is:

Under which conditions is the equalizer a normal subgroup?

This is true, for instance, if:

  • $\phi=\psi$ (in which case $C=G$);
  • $\phi(x)=\psi(x)$ implies $x=e$ (in which case $C=\{e\}$, the trivial subgroup);
  • Either $G$ or $H$ is abelian.

There must be other instances, though, and I imagine it is possible to give necessary and sufficient conditions on the homomorphisms for the equalizer to be normal.

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  • $\begingroup$ I'd say that the equality $\phi\circ\psi^{-1}(h)=\psi\circ\phi^{-1}(h)$ must hold true whenever the considered maps are well-defined. $\endgroup$
    – Sylvain JULIEN
    Dec 12, 2015 at 18:56
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    $\begingroup$ This is tautological, but isn't a necessary and sufficient condition that $\phi(x)^{-1} \phi(g) \phi(x) = \psi(x)^{-1} \phi(g) \psi(x)$ for all $x \in G$ and all $g \in C$. In other words, we need $\phi(x)\psi(x)^{-1} \in C_{H} (\phi(C))$ for all $x \in G$. $\endgroup$
    – Geoff Robinson
    Dec 12, 2015 at 20:17
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    $\begingroup$ Since every subgroup of a group $G$ is an equalizer subgroup (even in the category of finite groups), there will be very few conditions you can place in general to guarantee that the equalizer will be normal, and most of them will be in terms of severely restricting the types of morphisms, or the type of group (e.g., obviously it will hold for any Dedekind group). $\endgroup$
    – Arturo Magidin
    Dec 13, 2015 at 0:20

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