Why is this Isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$ 
Why is this table isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$?

Below is the additive composition table of four cosets of $\mathbb{Z}_2[x]/(x^2+x+1)$.
I don't understand why it's isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$. The reasoning should be really simple but I can't find why exactly..

 A: From having all zeroes in the diagonal, every nonzero element has order 2. So, if it is a group, it is $\mathbb Z_2\times\mathbb Z_2$.
Now you can start to think who can be $(1,0)$, $(0,1)$, and $(1,1)$.
A: Hint: map $(0,0) \mapsto 0$, $(1,0) \mapsto 1$, $(0,1) \mapsto x$, $(1,1) \mapsto 1+x$, can you show this is an isomorphism?
Another way to look at this: $x^2+x+1$ is irreducible over the field $\mathbb{F}_2$ and hence $\mathbb{F}_2[x]/(x^2+x+1) \cong \mathbb{F}_4$, which is a $\mathbb{F}_2$-vectorspace of dimension $2$. Hence the underlying abelian group structure is $\mathbb{Z}_2 \oplus \mathbb{Z}_2$.
A: Use your Cayley table and the one of $\Bbb Z_2\times \Bbb Z_2$ to prove that $0\mapsto (0,0)$, $1\mapsto (1,0)$, $x\mapsto (0,1)$, $1+x\mapsto (1,1)$ is indeed an isomorphism.
A: Rename $0=a,1=b,x=c,1+x=d$.
Adding $0$ induces the identity on $\{a,b,c,d\}$.
Adding $1$ induces the permutation $a\mapsto b,b\mapsto0,c\mapsto d,d\mapsto c$.
Adding $x$ induces the permutation $a\mapsto c,b\mapsto d,c\mapsto a,d\mapsto b$.
Adding $1+x$ induces the permutation $a\mapsto d,b\mapsto c,c\mapsto b,d\mapsto a$.
So, as a permutation group, it is $\{id,(ab)(cd),(ac)(bd),(ad)(bc)\}$.
