The standard definition of a **strict 2-group** says that it is a *strict monoidal category* in which every morphism is *invertible* and each object has a *strict inverse*.


Also it is a well known fact that a strict 2-group is a group object in the category of categories, and also a category object in the category of groups.


Now in terms of the latter two (equivalent) formulations of strict 2-groups, we must have a *group structure* on both the object set and the morphism set of the strict 2-group *induced by the associated bifunctor* of the strict monoidal category. But I am not able to understand how the standard definition actually implies that each *morphism* has an inverse with respect to this binary operation. Also I could not get how we are getting the associativity of this binary operation on the the morphism set.


I felt that from the standard definition we can only say that the *object* set is a group and that the *morphism* set is equipped with a *binary operation* with respect to which it has an *identity element*. (**NOTE**: we are talking about the binary operations on both object and morphism sets *induced by the associated bifunctor* of the strict monoidal category.) But, it seemed to me that existence of inverses and associativity are not guaranted.


Hence if those two equivalent formulations are true, and also the standard definition is correct, then I must be misunderstanding something here conceptually.


**Then where am I making the mistake?** 


If my question is stupid or not up to the standard of this forum then I apologize beforehand.


This is not a duplicate of the question [strict 2-groups VS crossed modules](https://mathoverflow.net/q/228804), because I felt in that question they discussed the invertibility of arrows in the categorical sense (that is, whether a group object in the category of categories has a groupoid structure or not); this fact is trivial and also mentioned in the paper by R. Brown and C. Spencer, G-groupoids, crossed modules and the fundamental groupoid of a topological group, Proc. Kon. Ned. Akad. v. Wet, 79, (1976), 296 – 302, [pdf]. Here I am asking about the existence of group inverses in the morphism set with respect to the binary operation induced from the *bifunctor of the strict 2-group* (*not* whether each morphism is invertible or not).


I hope my above explanation is sufficient to address the issue. If not, I will try to give more detail.


Thank you.



On the Group Structure of Morphism Set of a Strict 2-Group
If I understand the question correctly, then the answer for associativity is that the associator isomorphism in a non-strict monoidal category is natural, and that this naturality condition is still there in a strict monoidal category when the components of the associator are identity maps. Put differently, in a strict monoidal category, the two functors $C\times C\times C\to C$ don't just act equally on objects, they are equal *as functors*.


The construction of tensor-inverses for morphisms from the (*a priori* non-functorial) mere existence of tensor-inverses for objects is a bit trickier. Suppose $f:x\to y$, with composition-inverse $g:y\to x$ (so that $f\circ g = 1\_y$ and $g\circ f = 1\_x$); then the tensor-inverse of $f$ is the morphism


$$ x^{-1} = y^{-1} \otimes y \otimes x^{-1} \xrightarrow{1\otimes g \otimes 1} y^{-1} \otimes x \otimes x^{-1} = y^{-1}. $$


That this is a tensor-inverse of $f$ follows from a calculation involving naturality, most naturally (npi) done with string diagrams where the equalities such as $x \otimes x^{-1} = I$ are indicated by morphisms forming a dual pair (that just happen to be identity morphisms in this case).


