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consider the commutative diagram of group homomorphisms: $\begin{matrix} A&\stackrel{f}{\rightarrow}&B\\ \downarrow{g}&&\downarrow{k}\\ C&\stackrel{h}{\rightarrow}&D \end{matrix} $

suppose $B$, $C$ and $D$ are trivial groups $\{e\}$ does this imply that necessarely $A$ is also trivial?

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    No, $A$ can be anything you want and this still commutes.2011-06-07

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Note that the following answer is just an explanation of Qiaochu's answer in more concrete terms:

Let $x\in A$. We know that $k(f(x))=h(g(x))=e$ ($e$ is the identity of $D$) since the group $D$ is trivial (and, in particular, there can be only one choice for each of $k(f(x))$ and $h(g(x))$). Therefore, the diagram commutes if $D$ is trivial. In particular, we can take $A$ to be any non-trivial group and the diagram will commute irrespective of the choices of $B$ and $C$.

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    @Theo Buehler: thankyou very much that's what i was lookin$g$ $f$or!2011-06-07
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This diagram always commutes if $D$ is trivial, regardless of the other contents of the diagram. This is because the trivial group is the terminal object in the category of groups, and maps to the terminal object are unique.

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    ... the introduction to the Wikipedia article would look like a different language. But again, this is based only on my (limited) experience with people asking questions such as this one. Since you obviously have more experience on this forum compared to me, my comment probably should not be taken too seriously.2011-06-10