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Prove that there is a surjective homomorphism (an epimorphism) from $\Bbb Z*\Bbb Z$ onto $C_2*C_3$, where $A*B$ is the coproduct of $A$ and $B$ in $\mathsf{Grp}$.

Aluffi asks this question, oddly enough, immediately before revealing what $C_2*C_3$ actually is (which he does in the next exercise).

If I jump ahead and use the result of the next exercise, it's entirely straightforward. But without that, I can clearly see that there are surjective homomorphisms $f\colon \Bbb Z\to C_2$ and $g\colon \Bbb Z\to C_3$, so there must be a unique homomorphism $h\colon \Bbb Z*\Bbb Z\to C_2*C_3$ such that $h \circ i_{\Bbb Z_2}=i_{C_2}\circ f$ and $h\circ i_{\Bbb Z_3}=i_{C_3}\circ g$, but I see no obvious reason that this must be surjective.

Did Aluffi just mix up the order of the two exercises, or is there some other way to do it? Note: the previous exercise shows that there is a surjective homomorphism from $C_2 * C_3$ onto $S_3$, but I don't see how this could possibly be relevant.

dfeuer
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    Isn't $\Bbb Z * \Bbb Z$ just the free group on 2 symbols that maps onto any 2-generated group? – Myself Dec 11 '13 at 19:26
  • @Myself, that requires knowing that $C_2*C_3$ is generated by two elements, which doesn't show up till the next exercise. – dfeuer Dec 11 '13 at 19:31
  • I find it hard to work from universal definitions in this context. The question "can I add relations to $\langle x,y \mid ;\rangle$ to obtain $\langle x,y\mid x^2,y^3\rangle$" is comparatively trivial. (But depends on certain facts that you must have shown earlier.) – Myself Dec 11 '13 at 19:36
  • @Myself, facts I don't yet know, unfortunately. – dfeuer Dec 11 '13 at 19:40
  • It may be helpful to know whether the author, at this point in the text, has identified free groups with sets of words over an alphabet, or whether he restricts himself to universal properties? – Myself Dec 11 '13 at 19:42
  • @Myself, so far he has restricted himself to universal properties. I've read ahead a bit, so I have a limited understanding of what a free group is (I haven't done the exercises yet, so I don't really know), but no, not yet. – dfeuer Dec 11 '13 at 19:49
  • Has he defined the coproduct of groups in general at this point? Do you even know what $G\star H$ is for general groups? – Olivier Bégassat Dec 11 '13 at 19:51
  • @OlivierBégassat, no, I do not. I know only the universal property it satisfies. I'm guessing he just put these two exercises in the wrong order and I'm on a wild goose chase. – dfeuer Dec 11 '13 at 19:54
  • Well then you know what it is, it's the unique group up to isomorphism such that yada yada yada. From that definition alone you can at least work out that the canonical map $\Bbb Z\star\Bbb Z\to C_2\star C_3$ is an epimorphism without breaking a sweat, and epimorphisms are the same as surjective group homomorphisms. But the proof of that fact requires the tools you are currently learning I believe, so it's not entirely satisfying. – Olivier Bégassat Dec 11 '13 at 20:00
  • Please do write the exact place, No. of exercise, page in Aluffi's – DonAntonio Dec 11 '13 at 20:04
  • @DonAntonio, I don't have it with me right now. I will add that later. If you have the text, it's in the section of chapter II where he introduces the categories $\mathsf{Grp}$ and $\mathsf{Ab}$, right after the "Examples of Groups" section. – dfeuer Dec 11 '13 at 20:29
  • @OlivierBégassat, it's obvious that an onto homomorphism is epic, but no, he hasn't gotten to the converse yet. – dfeuer Dec 11 '13 at 20:30

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