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At n-lab (link), it says that

...the theory of groups lives naturally in the doctrine of categories with finite products, since a group object can be defined in any such category. Likewise, the theory of monoids lives in the doctrine of monoidal categories, and so on.

Why do we need finite products to define the notion of "group," but not "monoid"?

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    $\begingroup$ That's not entirely true. You could define "groups" within the doctrine of symmetric monoidal categories, but people prefer to call such a thing a Hopf algebra. $\endgroup$ – Zhen Lin Dec 27 '13 at 9:06
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    $\begingroup$ @Zhen: This is an answer. $\endgroup$ – Martin Brandenburg Dec 27 '13 at 10:33
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Well, to define 'group object' in a monoidal category $(\Bbb A,\otimes,I)$ we could start out from a monoid $(A,\mu,\eta)$ (i.e. $\mu:A\otimes A\to A,\ \eta:I\to A$ with usual associativity and unit expressing diagrams) with an additional inverse operation $()^{-1}:A\to A$. But.. what should this inverse operation satisfy? When we want to translate the equations $a\cdot a^{-1}=1$ and $\ a^{-1}\cdot a=1$, we would need a diagonal-like map $$\Delta: A\to A\otimes A$$ acting like "$a\mapsto(a,a)$", so the left hand side of the first equation which would act as "$a\mapsto a\cdot a^{-1}$" should be $\mu\circ \,\left({\rm id}_A\otimes()^{-1}\right)\,\circ\Delta$, and, by the way, for the right hand side "$a\mapsto 1$", we would need a map $\varepsilon:A\to I$ (which is automatically present and unique if $I$ is the terminal object). Thus, the first equation would translate to $$ \mu\circ \,\left({\rm id}_A\otimes()^{-1}\right)\,\circ\Delta = \nu\circ\varepsilon\,.$$ Both $\Delta$ and $\varepsilon$ are automatically present if $\otimes=\times$ (and thus $I$ is terminal).

However, as Zhen Lin commented, (if we assume a comonoid structure on $A$ with $\Delta$ and $\varepsilon$, then) the translations of the two equations for the inverse define the notion of Hopf monoid, where 'inverse' is renamed to 'antipode'.

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