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I am going to describe the diagram since I do not know how to do them on latex. Say $H$ is a subgroup of $G$ and we have $H\rightarrow G$ via inclusion (call this map $i$), and $H\rightarrow 1$ via the trivial map (call this map $z$). We want to find the pushout of it.

I think that the pushout will be $1$ because the map $f_1:G\rightarrow 1$ and the map $f_2:1\rightarrow 1$ are such that $f_2(z(h))=f_1(i(h))$ for all $h\in H$. Further, since $1$ is an initial object in the category of groups then we would be done.

Is the above correct?

Thanks.

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yes this seems correct because the square commutes and the universal property is easily proved by uniqueness of maps from (since it's a colimit) the zero object. Oh I called 1 the zero object, note it's initial and terminal in Grp. –  user51427 Dec 10 '12 at 2:18
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sorry I am wrong let me leave this comment up in case it is a common mistake. –  user51427 Dec 10 '12 at 2:40

1 Answer 1

up vote 3 down vote accepted

No, the trivial group is not the pushout. In the pushout diagram, $P$ is the pushout if whenever $Q$ is given such that $j_1\circ f = j_2 \circ g$, then there is a unique morphism $u$ that must also satisfy $j_1 = u \circ i_1$ and $j_2 = u \circ i_2$.

pushout

Thus, in your situation, the trivial group cannot be the pushout, since for any group $Q \neq 1$, with nontrival map $G \to Q$, that map cannot factor through the trivial map. Instead, what you're looking for is the amalgamated product. In your diagram, that would be the free product, $1 \ast G$, modulo the normal closure of $H$ (to account for the relations $i(h) = 1$ for $h \in H$). This amounts to $G/\langle H\rangle$, where $\langle H \rangle$ is the normal closure of $H$ in $G$.

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And in this case the colimit of the diagram coincides with the pushout, is that correct? –  Daniel Montealegre Dec 10 '12 at 5:07
    
Yes, the pushout is a type of colimit. –  Shaun Ault Dec 10 '12 at 12:06

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