Let $\mathcal{G}$ be a coherent 2-group. Following [Baez:HDA5](http://arxiv.org/abs/math/0307200) $\mathcal{G}$ is uniquely determined up to 2-equivalence by the following data: The fundamental group $\pi\_1(\mathcal{G})=G$, the second homotopy group $\pi\_2(\mathcal{G})=H$, the action of $G$ on $H$ by group autos and a class in $a\in H^3(G,H)$. Now theres a third group associated to $\mathcal{G}$ which consists of the morphisms of $\mathcal{G}$ and where multiplication is given by tensor product (horizontal composition in the 2-category). Of course this is only a monoid a priori but becomes a group after quotienting out the cumbersome to state but rather obvious congruence relation defined below.


Let $\epsilon\_X:I\rightarrow X\otimes\bar{X}$, $\eta\_X:\bar{X}\otimes X\rightarrow I$ be the unit and counit adjoint equivalences. If $f:X\rightarrow Y$ is an arrow in $\mathcal{G}$, there's the so called mate of $f$ denoted $\hat{f}:\bar{X}\rightarrow\bar{Y}$ and defined as 
$$\hat{f}=(i\_{\bar{X}}\otimes\epsilon\_Y)(i\_{\bar{X}}\otimes f^{-1}\otimes i\_{\bar{Y}})(\eta\_X\otimes i\_{\bar{Y}})$$
Here $i$ is the identity and I use composition from left to right! It is a result of Laplaza that for any object $X$ there's at most one isomorphism $X\cong I$ built out of $\eta$ and $\epsilon$. If there is one, call $X$ simple. Now define $f:X\rightarrow Y$ and $g:A\rightarrow B$ equivalent iff both $X\otimes\bar{A}$ and $Y\otimes\bar{B}$ are simple and the composition
$$I\rightarrow X\otimes\bar{A}\xrightarrow{f\otimes\hat{g}}{}Y\otimes\bar{B}\rightarrow I$$
is the identity.


Now from the above classification of 2-groups it should be possible to construct this group out of $G,H$ and $a$. Does anybody know how this operation on the data $G,H,a$ looks like? Is it a well known construction?



Groups lying horizontally in 2-groups
Let $\partial\colon C\_1\rightarrow C\_0$ be a crossed module, i.e. a group homomorphism together with a right action of $C\_0$ on $C\_1$, denoted exponentially $c\_1^{c\_0}$, satisfying the following laws:
$$\partial(c\_1^{c\_0})=-c\_0+c\_1+c\_0$$
$$c\_1^{\partial(c\_1')}=-c\_1'+c\_1+c\_1'.$$ 
Here I denote the group laws additively, despite they are non abelian in general.


Crossed modules are equivalent to strict $2$-groups. The strict $2$-group $C\_\*$ associated to $\partial\colon C\_1\rightarrow C\_0$ has set of objects in $C\_0$ and set of morphisms $C\_0\ltimes C\_1$, the semidirect product. I'll show that the group you define for $C\_\*$ is this semidirect product $C\_0\ltimes C\_1$. Hence it is **not invariant under tensor equivalences**.


The source and target of $(c\_0,c\_1)\in C\_0\ltimes C\_1$ are
$$(c\_0,c\_1)\colon c\_0+\partial(c\_1)\longrightarrow c\_0.$$
Composition is defined by $$(c\_0,c\_1)\circ(c\_0+\partial(c\_1),c\_1')=(c\_0,c\_1+c\_1').$$ 
The tensor product is the sum $+$ and the tensor unit is $0\in C\_0$. In this case $\epsilon$ and $\eta$ are the identity map on the tensor unit, i.e. $(0,0)\in C\_0\ltimes C\_1$. 


The mate of a morphism $f=(c\_0,c\_1)\in C\_0\ltimes C\_1$ is $\hat{f}=(-c\_0,-c\_1^{-c\_0})\in C\_0\ltimes C\_1$. The only simple object is the tensor unit $0\in C\_0$ and two morphisms $f=(c\_0,c\_1),g=(c\_0',c\_1')\in C\_0\ltimes C\_1$ are quivalent iff 
$$(0,0)=f+\hat{g}=(c\_0,c\_1)+(-c\_0',-(c\_1')^{-c\_0'})=(c\_0-c\_0',c\_1^{-c\_0'}-(c\_1')^{-c\_0'})
=(c\_0-c\_0',(c\_1-c\_1')^{-c\_0'})$$
i.e. iff $c\_0=c\_0'$ and $c\_1=c\_1'$, that is $f=g$. Therefore your group is simply the semidirect product $C\_0\ltimes C\_1$.


