# commutative diagrams of group homomorphisms

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. – Joe Johnson 126 Jun 7 '11 at 11:35

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 looking for! – palio Jun 7 '11 at 13:07

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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You might want to add that $A$ is trivial if and only if the diagram is a pull-back. – t.b. Jun 7 '11 at 11:46
Dear Qiaochu, do you think that this answer is appropriate for such a simple question? I do not mean to suggest that this is not a good answer; in fact, this point of view can be useful in a number of general situations. However, at least in my opinion, this is overkill for someone who is thinking about this kind of question. – Amitesh Datta Jun 7 '11 at 13:00
@Amitesh: I don't think this is overkill at all. In this context (exact sequences, commutative diagrams) it is essential to realize the (elementary) fact that there is precisely one map into and from $\{e\}$. This is used all the time. That category theorists have reserved a name (terminal/initial object) for such things, is just convenient, or at least worth knowing. – wildildildlife Jun 7 '11 at 13:55
@wildildildlife Every concept in mathematics is worth knowing. The point is not which concepts are worth knowing; rather which concepts are the most important to know in the context of this question. Qiaochu said that the reason the diagram commutes (when $D$ is trivial) is because the trivial group is the terminal object in the category of groups. Well, yes, it is but this is like describing a bear as a member of the family Ursidae. The OP would need to look up the notions of a category, of a morphism in a category, which are basic to some, but not to a person who cannot solve the question. – Amitesh Datta Jun 8 '11 at 0:09
@Amitesh: in the context of this question it is very important to know that $\{e\}$ is a terminal object. I agree with you that you don't need to call it that way, you don't need the category theoretic language. Perhaps you would have been happier if Qiaochu had added the phrase "in this case, it means that for every group $G$ there's precisely one arrow into $G\to \{e\}$." – wildildildlife Jun 8 '11 at 10:58