Exercise 6H on pp 45-46 of Herrlich & Strecker's Category theory (1979, 2d ed.) states (my emphasis):

(a) Show that if $K$ is a subgroup of the (finite) group $H$, then there exists a (finite) group $G$ and group homomorphisms $f_1, f_2:H\to G$ such that $$K = \{h\in H | f_1(h\,) = f_2(h\,)\}$$

[$\cdots$ lengthy hint on how to construct $(G, f_1, f_2)$ omitted $\cdots$]

(b) Use part (a) to show that the epimorphims in Grp are precisely the surjective homomorphisms; likewise in the category of finite groups.

My question is:

why do H&S take the trouble of making a special case out of the category of finite groups?

The emphasized text in the quoted exercise, especially the last phrase of part (b), suggest that the proof of the "epic $\Leftrightarrow$ surjective" equivalence for the category Grp somehow does not hold for the category of finite groups. How could this be? After all, finite groups are groups too, so any theorem that is proved for groups in general would necessarily hold for finite groups as well. It would have made sense if the proof for the case of finite groups did not apply for groups in general, but not the other way around.

(FWIW, after doing parts (a) and (b) for Grp, I don't see any step in the argument that depends on the cardinality of the groups involved. In particular, this is the case for the proposed construction given in the omitted hint.)

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    $\begingroup$ The argument may be essentially the same for both categories, but the result for FinGrp isn’t an immediate corollary of the result for Grp: you need to know that the group $G$ constructed in (a) is finite if $H$ is. That is, the result for FinGrp may be a corollary of the proof of the result for Grp, but it doesn’t follow from the result for Grp itself. $\endgroup$ Commented Dec 26, 2011 at 21:21
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    $\begingroup$ @BrianM.Scott OK, but that reasoning just raises the question: why not add similar riders of the form "likewise for the category of countable groups, and the category of torsion groups, and the category of Lie groups, etc., etc."? What is it about finite groups in particular that merits the special mention? $\endgroup$
    – kjo
    Commented Dec 26, 2011 at 21:35
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    $\begingroup$ There is some history here. The first generation proof that epis were surjections in Grp proceeded by using the free product construction, which doesn't work in the finite group case. The more modern proof of using permutations came later, by Linderholm (I think) in the 1970s. $\endgroup$
    – user641
    Commented Dec 26, 2011 at 21:58
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    $\begingroup$ @Steve: Yes: Linderholm, Carl (1970). A Group Epimorphism is Surjective. American Mathematical Monthly 77, pp. 176–177. The argument is summarized in this sci.math post by Arturo Magadin. $\endgroup$ Commented Dec 26, 2011 at 23:24
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    $\begingroup$ Basic group theory courses have a lot of more important material to cover. It’s not clear to me that this construction is useful for much of anything of a purely group-theoretic nature (i.e., outside the category-theoretic setting). $\endgroup$ Commented Dec 27, 2011 at 1:34

1 Answer 1


Neither part (a) nor part (b) for finite/countable groups follow from the case of general groups.

In general, if $\mathbf{D}$ is a (full) subcategory of $\mathbf{C}$, then:

  1. If $A,B\in\mathbf{D}$ and $f\colon A\to B$ is an epimorphism in $\mathbf{C}$, then it is an epimorphism in $\mathbf{D}$. However, if $f$ is an epimorphism in $\mathbf{D}$, it may or may not be an epimorphism in $\mathbf{C}$.

    The reason is that if the morphism is cancellable on the right when composed with any arrows in $\mathbf{C}$, then it is certainly cancellable on the right when composed with arrows in $\mathbf{D}$. One may view the fact that group homomorphisms that are surjective on the underlying sets are epimorphisms in categories of groups/rings/etc. as a special case of this, since surjective maps are epimorphisms in $\mathbf{Sets}$.

    However, it is possible for a map to be cancellable on the right when composed with arrows in $\mathbf{D}$, but not when composed with arrows in $\mathbf{C}$ (fewer arrows to compose with). Even when the subcategory is full. For example, the variety of groups generated by $A_n$, $n\geq 5$, is a full subcategory of the category of groups (in fact, it's a reflective subcategory). The embedding $A_{n-1}\to A_n$ is an epimorphism in this variety, but not in the category of all groups (or of all finite groups, or even in the slightly larger category given by the variety generated by $A_{n+1}$).

  2. A similar thing can happen with equalizer subobjects. Part 1 of the exercise shows that every subgroup is an equalizer subgroup (a subgroup that arises as the equalizer of a pair of morphisms). If $K$ is an equalizer subobject of $H\in\mathbf{D}$, then it is an equalizer subobject in the larger category (same witnesses work), but it is possible for a subobject to be an equalizer subobject in the larger category but not in the smaller one. T

    The statement that all subgroups are equalizer subgroups is stronger than the statement that epimorphisms are surjective. In the variety of nilpotent groups of class at most $2$ (or the category of finite groups of nilpotency class at most $2$), every epimorphism is surjective, but not every subgroup is an equalizer subgroup (the deviation of a subgroup from being an equalizer subgroup is measured by the dominion of the subgroup relative to the variety; see here for the basic references, here for a brief discussion, and here for a somewhat off-topic semi-extensive discussion).

As noted in comments, the fact that this holds for finite groups as a special case is usually highlighted because Schreier himself did so in his paper on the amalgamated product, though his proof that any finite amalgam can be embedded in a finite group is much more complicated (and I don't think Linderholm's construction can be easily adapted for the more general statement, when we have a single group $K$, embeddings into two subgroups $H_1$ and $H_2$, and then embeddings from $H_1$ and $H_2$ into a third group $G$ in such a way that the two $H_i$ intersect, as subgroups, precisely at $K$, and that the induced embeddings of $K$ into $G$ agree). But there are lots of natural subcategories (nilpotent groups of class at most $n$, finitely generated nilpotent groups of any class, solvable groups, varieties, etc) where your (1) may fail even if (2) holds, or (2) may fail.


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