# 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?

• Have you tried to verify $G$ is abelian? This might simplify the overall verification. – hardmath Sep 29 '16 at 16:42
• Hint: what is the group identity of $(G,\cdot_f)$? That element must be mapped to the group identity of $E\times K$ (under any group homomorphism, so certainly under an isomorphism). And it seems that the first coordinate already acts the way you want, so the sought isomorphism is probably pretty simple there.... – Greg Martin Sep 29 '16 at 16:49
• @hardmath Ah nice. I did not observe that until now. Thanks. I will consider this. – TheGeekGreek Sep 29 '16 at 18:52
• @GregMartin The group identity is given by $(0,-\varphi_f(0,0)) = (0,f(0))$. Yes, the first coordinate is alright. So we would like that the isomorphism maps $(0,f(0))$ to $(0,0)$. So I would say $\iota(e,k) := (e,k-f(0))$. – TheGeekGreek Sep 29 '16 at 18:54
• Great, so we have a concrete map in mind; can you determine whether that map is an isomorphism? – Greg Martin Sep 29 '16 at 22:20

## 1 Answer

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 $\psi(e,k)= (e,k-f(e))$.