$\mathbf{B}A$ as a 2-group in a long fiber sequence

Question

I am trying to digest the following statement about 2-group:

From nlab Observation 4.2:

"Let $A \to \hat G$ be the inclusion of a subgroup, exhibiting a central extension $A \to \hat G \to G$ with $G := \hat G/A$. Then this short exact sequence of groups extends to a long fiber sequence of 2-groups $$ A \to \hat G \to G \to \mathbf{B}A \to \mathbf{B}\hat G \to \mathbf{B}G \to \mathbf{B}^2 A \,, $$ where $\mathbf{B}A$ denotes the 2-group given by the crossed module $(A \to 1)$, and similarly for the other cases."

Question 1: I understand the induced long fiber sequence. But I do not understand why $\mathbf{B}A$ is called a 2-group? Can someone help me on this? Also what is the emphasis on "given by the crossed module $(A \to 1)$"?

Here the connecting homomorphism $G \to \mathbf{B}A$ is presented in the category of crossed modules by a zig-zag / anafunctor whose left leg is the above weak equivalence:

Question 2: Am I correct that to discuss 2-group $\mathbf K$ in this context we need to have the classifying space $\mathbf{B}K$ such that both the homotopy groups of $\pi_1(\mathbf{BK}) \neq 0$ and $\pi_2(\mathbf{BK}) \neq 0$? For example, if $\mathbf K=K_1 \times \mathbf{B}K_2$, then $$\pi_1(\mathbf{BK}) = \pi_1(\mathbf{B}K_1) = K_1$$ and $$\pi_2(\mathbf{BK})= \pi_2(\mathbf{B}^2K_2) = K_2.$$

Are these two conditions sufficient conditions?

Or are these two conditions only necessary conditions?

Answer 1

A 2-group is a groupoid equipped with a monoidal structure in which every object is invertible. For a 1-object groupoid $BG$, the Eckmann-Hilton argument shows that $BG$ admits a unique strictly monoidal 2-group structure if and only if $G$ is abelian. The reference to crossed modules is because the latter are equivalent to 2-groups.

Your second question is a bit unclear, but if $K_2$ is abelian then there is indeed a 2-group $K_1\times BK_2$, a product of strictly monoidal groupoids.