Vtyjkyuk

We define a group law on the set
$$
Q(K) := mathbbZ/2mathbbZ times K^bullet/K^bullet 2
$$
as follows
beginalign*
(0, alpha) + (0, beta) &= (0, alpha beta) \
(1, alpha) + (0, beta) &= (1, alpha beta) \
(1, alpha) + (1, beta) &= (0, -alpha beta).
endalign*
It is trivial to check that, in fact, this satisfies the axioms of an abelian group.
$Q(K)$ is called the extended square class group.
Next, one checks equally trivially that
$$
(e,d) colon W(K) to Q(K)
$$
is a group homomorphism.
Thus, $e$ and $d$ together yield a reasonable invariant.

(Original image here.)

This is excerpt from W. Scharlau: Quadratic and Hermitian Forms,
page 37.

In this $Q(K)$ we define group law as given I don't understand what is third law saying $(1,alpha) + (1,beta) = (0,-alphabeta)$ is fine in first coordinate since it is from two element group but why $-alpha beta$ in second coordinate. Also how it satisfy abelian axioms? What I think is since both groups are abelian there cartesian product is also abelian. And last question is how to show it is homomorphisam?

Let $G$ be a group containing an element $g$ of the center of $G$.
Let us define a law on the set $K = mathbbZ/2mathbbZ times G$ by setting
beginalign
(0,x)(0,y) &= (0,xy) \
(0,x)(1,y) &= (1,xy) \
(1,x)(0,y) &= (1,xy) \
(1,x)(1,y) &= (0,gxy)
endalign
I claim that $K$ is a group for this law. To check associativity, one needs to compare
$A = bigl((a,x)(b,y)bigr)(c,z)$ and $B = (a,x)bigl((b,y)(c,z)bigr)$. The only nontrivial case occur when at least two elements of the set $a, b, c$ are equal to $1$. Now,
beginalign
bigl((1,x)(1,y)bigr)(0,z) &= (0,gxy)(0,z) = (0,gxyz) = (1,x)(1,yz)= (1,x)bigl((1,y)(0,z)bigr) \
bigl((1,x)(0,y)bigr)(1,z) &= (1,xy)(1,z) = (0,gxyz) = (1,x)(1,yz)= (1,x)bigl((0,y)(1,z)bigr) \
bigl((0,x)(1,y)bigr)(1,z) &= (1,xy)(1,z) = (0,gxyz) = (0,x)(1,yz)= (0,x)bigl((1,y)(1,z)bigr) \
bigl((1,x)(1,y)bigr)(1,z) &= (1,gxy)(1,z) = (1,gxyz) = (1,x)(1,yz)= (1,x)bigl((1,y)(1,z)bigr)
endalign
Clearly $(0,1)$ is the identity element, the inverse of $(0,x)$ is $(0, x^-1)$ and the inverse of $(1,x)$ is $(1,g^-1x^-1)$. Thus $K$ is a group.

For your question, just take $g = -1$.