Tensor product of groups maps to both groups

Statement

Suppose $G$ and $H$ are groups with a compatible pair of actions $\alpha :G\to \operatorname {Aut} (H)$ and $\beta :H\to \operatorname {Aut} (G)$ on each other. Suppose $G\otimes H$ is the tensor product of groups for this compatible pair of actions. Then, there are natural homomorphisms:

$\lambda _{G}:G\otimes H\to G,\lambda _{H}:G\otimes H\to H$

given by extending the following mapping from the generating set:

$\lambda _{G}(g\otimes h)=g(\beta (h)(g^{-1}))$

$\lambda _{H}(g\otimes h)=(\alpha (g)(h))h^{-1}$

If we use $\cdot$ to denote both the actions $\alpha$ and $\beta$, then this becomes:

$\lambda _{G}(g\otimes h)=g(h\cdot g^{-1})$

$\lambda _{H}(g\otimes h)=(g\cdot h)h^{-1}$

In particular, combining these, we get a natural homomorphism to the external direct product:

$G\otimes H\to G\times H$

Particular cases

• In case that both $G$ and $H$ are abelian groups and the actions are trivial, both the homomorphisms are trivial, hence the homomorphism to the direct product is also trivial.
• In case that $G=H$ and we are using the action by conjugation, so that we are mapping from the tensor square of a group, both maps are identical to each other, and they are both the same as the commutator map viewed as a homomorphism (which factors through the exterior square).
• More generally, if they are both subgroups inside some big group and normalize each other, then both mappings are equal and are given by a commutator mapping in the big group.

Proof

We will give below the proof for the map $G\otimes H\to G$. The proof for the map $G\otimes H\to H$ is perfectly analogous.

To do the proof, we need to demonstrate that the set map:

$G\times H\to G$

given by:

$(g,h)\mapsto g(\beta (h))g^{-1}$

is a crossed pairing for the compatible pair of actions $\alpha ,\beta$.

The checking of all the conditions is relatively straightforward.

Checking the condition with respect to $\alpha$ (the action of $G$ on $H$): framed the easy way using the $\cdot$

The condition for a general function $f$ would read:

$f(g_{1}g_{2},h)=f(g_{1}\cdot g_{2},g_{1}\cdot h)f(g_{1},h)$ for all $g_{1},g_{2}\in G,h\in H$

For our particular function, what we need to show is below:

To prove: $(g_{1}g_{2})(h\cdot (g_{1}g_{2})^{-1})=(g_{1}\cdot g_{2})((g_{1}\cdot h)\cdot (g_{1}\cdot g_{2}^{-1}))(g_{1})(h\cdot g_{1}^{-1})$

Proof: We start from the right side and perform

Step no.	Current status of right side	Facts/properties used	Specific rewrites
1	$(g_{1}\cdot g_{2})(g_{1}\cdot (h\cdot g_{2}^{-1}))(g_{1})(h\cdot g_{1}^{-1})$	Compatibility of the actions	Replace $(g_{1}\cdot h)\cdot (g_{1}\cdot g_{2}^{-1})$ by $g_{1}\cdot (h\cdot g_{2}^{-1})$
2	$g_{1}\cdot ((g_{2})(h\cdot g_{2}^{-1})))(g_{1})(h\cdot g_{1}^{-1})$	Use that conjugation by $g_{1}$ is a group action of $G$ on itself	Replace $(g_{1}\cdot g_{2})(g_{1}\cdot (h\cdot g_{2}^{-1}))$ by $g_{1}\cdot ((g_{2})(h\cdot g_{2}^{-1}))$.
3	$g_{1}(g_{2}(h\cdot g_{2}^{-1}))g_{1}^{-1}g_{1}(h\cdot g_{1}^{-1})$	Explicitly write the conjugation by $g_{1}$ in terms of group multiplication	Replace $g_{1}\cdot ((g_{2})(h\cdot g_{2}^{-1}))$ by $g_{1}(g_{2}(h\cdot g_{2}^{-1}))g_{1}^{-1}$
4	$g_{1}g_{2}(h\cdot g_{2}^{-1})(h\cdot g_{1}^{-1})$	Cancel the $g_{1}^{-1}$ and $g_{1}$ appearing next to each other	Replace $g_{1}^{-1}g_{1}$ by the identity (empty word)
5	$g_{1}g_{2}(h\cdot (g_{2}^{-1}g_{1}^{-1}))$	Use that the action of $H$ on $G$ is a group action	Rewrite $(h\cdot g_{2}^{-1})(h\cdot g_{1}^{-1})$ as $h\cdot (g_{2}^{-1}g_{1}^{-1})$.
6	$(g_{1}g_{2})(h\cdot (g_{1}g_{2})^{-1})$	Use that inverse map is involutive	Replace $g_{2}^{-1}g_{1}^{-1}$ by $(g_{1}g_{2})^{-1}$, also put parentheses around beginning $g_{1}g_{2}$.

Checking the condition with respect to $\beta$ (the action of $H$ on $G$): framed the easy way using the $\cdot$

The condition for a general function $f$ to satisfy this are:

$f(g,h_{1}h_{2})=f(g,h_{1})f(h_{1}\cdot g,h_{1}\cdot h_{2})$ for all $g\in G,h_{1},h_{2}\in H$

For our particular function, what we need to prove is:

To prove: $g((h_{1}h_{2})\cdot g^{-1})=g(h_{1}\cdot g^{-1})((h_{1}\cdot g)((h_{1}\cdot h_{2})\cdot (h_{1}\cdot g)^{-1}))$

Proof: We will simplify the right side to the left side.

Step no.	Current status of right side	Facts/properties used	Specific rewrites
1	$g((h_{1}\cdot g^{-1})(h_{1}\cdot g))((h_{1}\cdot h_{2})\cdot (h_{1}\cdot g)^{-1}))$	associativity of group multiplication	just a re-parenthesization, no rewrites
2	$g(h_{1}\cdot (g^{-1}g))((h_{1}\cdot h_{2})\cdot (h_{1}\cdot g)^{-1}))$	The action of $H$ on $G$ is a group action	Rewrite $(h_{1}\cdot g^{-1})(h_{1}\cdot g)$ as $h_{1}\cdot (g^{-1}g)$
3	$g((h_{1}\cdot h_{2})\cdot (h_{1}\cdot g)^{-1}))$	We simplify $g^{-1}g$ to the identity element and hence also simplify $h_{1}\cdot (g^{-1}g)$ to the identity element.	Rewrite $h_{1}\cdot (g^{-1}g)$ as the empty word (the identity element).
4	$g((h_{1}\cdot h_{2})\cdot (h_{1}\cdot g^{-1})$	We use that the action of $H$ on $G$ is an action and hence preserves inverses.	Rewrite $(h_{1}\cdot g)^{-1}$ as $h_{1}\cdot g^{-1}$.
5	$g(h_{1}h_{2}h_{1}^{-1}\cdot (h_{1}\cdot g^{-1}))$	Expand out the meaning of conjugation	Rewrite $h_{1}\cdot h_{2}$ as $h_{1}h_{2}h_{1}^{-1}$
6	$g((h_{1}h_{2}h_{1}^{-1}h_{1})\cdot g^{-1})$	Use that the action is a group action	Rewrite $h_{1}h_{2}h_{1}^{-1}\cdot (h_{1}\cdot g^{-1})$ as $(h_{1}h_{2}h_{1}^{-1}h_{1})\cdot g^{-1}$
7	$g((h_{1}h_{2})\cdot g^{-1})$	Simplify a group product	Rewrite $h_{1}h_{2}h_{1}^{-1}h_{1}$ as $h_{1}h_{2}$ by canceling the $h_{1}^{-1}$ and $h_{1}$ that appear adjacent to each other.