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**Deveno** Recall that $(a,b)N = (a',b')N$ if and only if $(a,b)\ast[(a',b')]^{-1} \in N$.

Now:

$(a,b)\ast[(a',b')]^{-1} = (a,b)\ast(a'^{-1},b'^{-1}) = (aa'^{-1},bb'^{-1})$.

Since $A$ is a group, we certainly have $aa'^{-1} \in A$, so $(aa'^{-1},bb'^{-1}) \in N = A \times \{1_B\}$ if and only if $bb'^{-1} = 1_B$, and thus $b = b'$.

Again, why are you proving the function is well defined?

It should be clear that $(a,b)N \mapsto b$ is a homomorphism (it's a straightforward computation).

Homomorphism, got it

Also $(1_A,b)N$ is clearly a pre-image for $b$.

Surjective, got it

So all you need to do is show $\text{ker }\psi = N$ (hint: $(a,1_B)N = N$).

Sorry, what does this do? Can you be more explicit, like say what you are trying to show at each step?