Free algebra identity (empty word) and involution

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I didn't find the way to define an involution on the free algebra (say in two generators) yet. Specifically I want unitary elements. Thus, I added two variables, $y$ and $z$ which serve as $w^\star,x^\star$ and force a relation so that $w y=1$, the empty word, and similar relation $x z=1$ . But then typing

> A.<w,x,y,z> = FreeAlgebra(QQ,4)  
> G=A.g_algebra({w*y: 1})    
(x,y,z) =
> G.gens()

yields an error (#This is dirty, coercion is broken). That is: Question: which is the free algebra empty word here?

Of course, I tried to add a symbol $E$ which then satisfies $gE=g=Eg$ for all the generators $g$, But then

> G=A.g_algebra({w*y: E})

yields also an error.

> /usr/lib/python3/dist-packages/sage/algebras/free_algebra.py
> in g_algebra(self, relations, names,    
> order, check)    
>     873                     d_poly = commuted - self(c) * self(m)     
>     874                     break    
> --> 875             assert c_coef is not None, list(m)   
>     876             v2_ind = self.gens().index(v2)   
>     877             v1_ind = self.gens().index(v1)   
> 
> AssertionError: [(E, 1)]

Second question: does somebody know an easier way to to implement the involution here?