Show that two certain groups are isomorphic Let $E,K$ be two abelian groups (additively written) and let the group law of $(G,\cdot_f)$ be defined as $$(e,k)\cdot_f(e',k') = (e + e',k + k'+ \varphi_f(e,e'))$$
where $$\varphi_f(e,e') = f(e+e') - f(e) - f(e')$$ for some function $f:E \to K$. We have the identity $$\varphi_f(e + e',e'') + \varphi_f(e,e') = \varphi_f(e,e'+e'') + \varphi_f(e',e'')$$ for any $e,e',e''\in E$ and thus $$\varphi_f(0,0)=\varphi_f(e,0)=\varphi_f(0,e')$$ Now I should show, that $(G,\cdot_f)$ is isomorphic to $E \times K$ with usual componentwise operation. My thought was to construct an isomorphism, but this did not guide me to an answer. Is there another way or a more strategic way of finding such an isomorphism?
 A: Method 1: Guess $\psi : G\to E\times K$ should be defined by $\psi(e,k) =(e,k-f(e))$ and check it is an isomorphism.
Method 2: Guess that $\psi : G\to E\times K$ should be of the form $\psi(e,k)=(e,k+\theta(e))$. Then since we want $\psi$ to be a homomorphism we obtain the equation:
$$(e+e',k+k'+\phi_f(e,e')+\theta(e+e'))=(e+e',k+k'+\theta(e)+\theta(e'))$$
and hence
$$\theta(e+e')=\theta(e)+\theta(e')-\varphi_f(e,e')=\theta(e) + f(e) + \theta(e') + f(e') - f(e+e')$$
so that
$$\theta(e+e')+f(e+e')=\theta(e) + f(e) + \theta(e') + f(e').$$
From this we see that if $\theta(e)=-f(e)$, then $\psi$ will be a homomorphism. Check it is an isomorphism.
Method 3:
Much too long and far to complicated but more structured.

Theorem: Let $K, E$ and $G$ be groups and let $p:G\to E$, $s:E\to G$, $q:G\to K$ and $r: K\to G$ be group homomorphisms such that $p\circ s=1_E$, $qr=1_K$, $pr=0$ and $qs=0$. If $G$ is abelian and $rq+sp=1_G$, then the map $\psi : G\to E\times K$ defined by $\psi(g) = (p(g),q(g))$ is an isomorphism.
Proof:
$\psi$ is a bijection since $\theta$ defined by $\theta(e,k) = s(e) + r(k)$
is its inverse. $$\theta(\psi(g))= \theta(p(g),q(g))=s(p(g))+r(q(g))=(rq+sp)(g)=g$$
and
$$\psi(\theta(e,k))=(p(s(e)+r(k)),q(s(e)+r(k))=(p(s(e))+p(r(k)),q(s(e))+q(r(k)))=(e+0,0+k)=(e,k)$$
Finally $\psi$ is a canonical map into a product (and hence a homomorphism). More concretely
$$\psi(g_1+g_2) = (p(g_1)+p(g_2),q(g_1)+q(g_2))=(p(g_1),q(g_1))+(p(g_2),q(g_2))=\psi(g_1)+\psi(g_2).$$

Now back to the problem. Notice that there is a natural homomorphism $p: G\to E$ defined by $p(e,k)=e$. This map has a section $s: E\to G$ which is also a homomorphism defined by $s(e)=(e,f(e))$. The kernel of $p$ is $\{(0,k)\,|\,k \in K\}$. Let us write $r : K \to G$ for the homomorphism defined by $r(k) = (0,k+f(0))$ which is an isomorphism onto the kernel of $p$. Since $G$ is abelian it follows that the map defined by $u(e,k) = (e,k)\cdot_f s(p(e,k))^{-1}$ is a homomorphism. Now since as mentioned $r$ is an isomorphism onto the kernel of $p$ and $$p(u(e,k))= p((e,k)\cdot_f s(p(e,k))^{-1}) = p(e,k) -   p(s(p(e,k))) = e-e=0$$ it follows that $u$ factors through $r$ via a homorphism $q: G\to K$. The theorem above then tells us that desired isomorphism will be $\psi : G\to E\times K$ defined by $$\psi(e,k)=(p(e,k),q(e,k))=(e,q(e,k)).$$ Now we would like an explicit formula $\psi$ and hence for $q$ and hence also for $u$. Since $s(p(e,k))=(e,f(e))$, solving $$(0,f(0)) = (e,f(e)) \cdot_f (e',k')= (e+e',f(e)+ k' + \varphi(e,e'))=(e+e',f(e)+k'+f(e+e')-f(e)-f(e'))=(e+e',k'+f(e+e')-f(e'))$$ for $e'$ and $k'$ we see that $e'=-e$ and hence $k' = f(0) - f(e -e)+f(-e)=f(-e)$ and hence $$u(e,k) = (e,k)\cdot_f (-e,f(-e)) = (0,k + f(-e) + \varphi(e,-e))=(0,k+f(-e)+f(0)-f(e)-f(-e))=(0,k + f(0)-f(e)).$$ Now since $rq=u$ solving for $q(e,k)$ in the equation
$r(q(e,k))=u(e,k)$ we find that $q(e,k)=k-f(e)$ and hence $\psi(e,k)= (e,k-f(e))$.
