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Let $G$ be any group. Form a category $\mathcal{G}$ as follows: object is only one, say *, and morphisms from $*$ to $*$ are the elements of $G$, and composition of morphisms is product of elements of $G$.

Here, the number of morphisms from $*$ to $*$ is equal to $|G|$.

I was trying to see whether co-product exists in this category, and I was feeling that it does not exist, if $|G|$ is at least $2$.

Now by co-product of $*$ with $*$, we mean an object (which should be $*$ only), with a pair of morphisms $x:*\rightarrow *$ and $y:*\rightarrow *$ with following universal property:

if $a:*\rightarrow *$ and $b:*\rightarrow *$ is a pair of morphisms, there exists a unique morphism $\theta:*\rightarrow *$ such that $$x=\theta a \mbox{ and } y=\theta b.$$

Now $|G|$ is at least $2$, so (i) If morphisms $x$ and $y$ are same, then consider $a$ and $b$ distinct, last equation gives contradiction. (ii) If morphisms $x$ and $y$ are distinct, choose morphisms $a$ and $b$ to be same, again contradiction. So co-product of $*$ with $*$ does not exists in $\mathcal{G}$.

Is this justification correct? If yes, then considering opposite category, we can conclude that product does not exist in this category. Am I correct?

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    $\begingroup$ Yes to all questions. $\endgroup$ – Ittay Weiss Sep 24 '16 at 3:58
  • $\begingroup$ Thanks for convincing. $\endgroup$ – p Groups Sep 24 '16 at 7:40

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