Say a sequence $A\overset{f}{\to}B\overset{g}{\to}C$ of group morphisms is coexact if $\operatorname{Coker}f= \operatorname{Im}g$.

The first isomorphism theorem ensures that exactness implies coexactness. In an abelian category they coincide (because the property of being abelian is autodual). Where do they differ?

Question. What are some examples of coexact sequences of groups morphisms which are not exact?

I think we want the normal closure of $\operatorname{Im}(f)\leq B$ to be strictly smaller than $\operatorname{Ker}g$, but I'm really bad with all this normal stuff.

  • $\begingroup$ I'm not sure how to interpret $\operatorname{Coker}f= \operatorname{Im}g$ other than the fact that $g$ factors through $\operatorname{Coker}f$ and gives an isomorphism with the image, which is just a reformulation of exactness. $\endgroup$ – Captain Lama Jun 1 '19 at 1:12
  • $\begingroup$ @CaptainLama: But if the image of $f$ is not normal, the cokernel of $f$ is not just the quotient by the image of $f$! $\endgroup$ – Eric Wofsey Jun 1 '19 at 2:02
  • $\begingroup$ @EricWofsey Of course, silly me... $\endgroup$ – Captain Lama Jun 1 '19 at 18:34

Let $A$ be a non-normal subgroup of $B$ and let $N$ be the normal subgroup it generates. Then the sequence $A\to B\to B/N$ is coexact: the cokernel of $A\to B$ is the universal morphism out of $B$ that annihilates $A$, which is just the quotient $B\to B/N$.

More generally, coexactness of $A\overset{f}{\to}B\overset{g}{\to}C$ means precisely that the image of $f$ generates the kernel of $g$ as a normal subgroup, since the cokernel of $f$ is exactly the quotient by normal subgroup generated by its image.

  • $\begingroup$ Dear Eric, this is what I meant in the last sentence of my question. What I tried to ask for is an explicit example :) $\endgroup$ – Arrow Jun 1 '19 at 8:24
  • $\begingroup$ @Arrow As an explicit example you can take $B = D_8$ = dihedral group of order eight which has the presentation $\langle x,a \mid a^4 = x^2 = e, xax^{-1} = a^{-1} \rangle$. Then the subgroup $\{e, x \}$ of order $2$ is non-normal. Its normal closure is $\{ e,x,a^2,a^2x \}$. See groupprops.subwiki.org/wiki/Dihedral_group:D8. $\endgroup$ – Paul Frost Jun 2 '19 at 16:49

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