# Exact sequence giving kernel of mapping from tensor square to exterior square

## Statement

Let $G$ be any group. There is a natural exact sequence as follows:

$H_3(G;\mathbb{Z}) \to \Gamma(G^{\operatorname{ab}}) \to G \otimes G \to G \wedge G \to 0$

Here:

• $H_3(G;\mathbb{Z})$ denotes the third homology group for trivial group action of $G$ on $\mathbb{Z}$.
• $\Gamma$ denotes the universal quadratic functor and $G^{\operatorname{ab}}$ denotes the abelianization of $G$.
• $G \otimes G$ denotes the tensor square of $G$, i.e., the tensor product of $G$ with itself where both parts of the compatible pair of actions are the action of $G$ on itself by conjugation.
• $G \wedge G$ is the exterior square of $G$.

The maps are as follows:

• The first map is mysterious (some kind of connecting homomorphism?)
• The second map, $\Gamma(G^{\operatorname{ab}}) \to G \otimes G$, sends PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]
• The third map, $G \otimes G \to G \wedge G$, is the natural quotient map that sends a generator of the form $x \otimes y$ to the corresponding element $x \wedge y$.