Let $G$ be an abelian group and $\phi :G\to H$ is a surjective homomorphism with kernel $K$. Suppose there is a homomorphism $\psi :H\to G $ such that $\phi\psi$ is identity map on $H$. Show $K\times H\cong G$

This is a bit like first homomorphism theorem , we already know $G/K\cong H$. What confuse me is to define a proper isomorphism.

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    Map $(k,h)\mapsto k\psi(h)$ for all $k\in K, h\in H$, and check everything that needs to be checked. – Jyrki Lahtonen Dec 6 at 14:18
  • You have a canonical inclusion map $K\hookrightarrow G$, and you're given a homomorphism $\psi:H\to G$. By definition of $\times$, this gives you a homomorphism $K\times H\to G$ (a priori this is not necessarily the same homomorphism as Jyrki's above, but it will turn out to be). Check everything that needs to be checked. – Arthur Dec 6 at 14:20
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    Take $g\in G$. Let $h = \phi(g)$ and $k = g\psi(h)^{-1}$, which makes $k\psi(h) = g$. You need to show that $k\in K$, and here I suspect it is crucial that $K$ is the kernel of $\phi$. – Arthur Dec 6 at 14:46
  • That's an awesome statement! I'm wondering where can we use this result? – mathnoob Dec 6 at 15:05
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    @mathnoob This result is called the splitting lemma and since it has a name, it's important. – Arthur Dec 6 at 15:06

So consider a map $f :K \times H \rightarrow G$, $f((k,h))=k+\psi(h)$. Now we show that this map is an isomorphism: $f((k_1,h_1)+(k_2,h_2))=f((k_1+k_2,h_1+h_2))=(k_1+k_2)+\psi(h_1+h_2)=(k_1+k_2)+(\psi(h_1)+\psi(h_2))=k_1+\psi(h_1)+k_2+\psi(h_2)=f((k_1,h_1))+f((k_2+h_2))$. So this show $f$ is an homomorphism.

Now for bijectivity, Consider $f((k_1,h_1))=f((k_2,h_2))$, this means $k_1+\psi(h_1)=k_2+\psi(h_2)$. Apply $\phi$ to both side of the equation to get $\phi(\psi(h_1))=\phi(\psi(h_2))$ which implies $h_1=h_2$ which implies $k_1+h_1=k_2+h_2$ which implies $k_1=k_2$ which implies $(k_1,h_1)=(k_2,h_2)$. So this shows injectivity.

Now for surjectivity, Let $g\in G$, $k=g\psi(\phi(g))^{-1}$, then $\phi(k)=\phi(g)\phi(g)^{-1}=1$ so $k\in K$ and then $g=k\psi(\phi(g))$. So $(k, \phi(g))$ maps to $g$.

It is a well known result that if $K, L$ are normal in some group $G$ and $KL = G$, $K \cap L = 1$, then $G \simeq K \times L$.

From the fact that $\phi\psi = 1_H$, we know that $\psi$ is injective (Hint: take elements that map to the same image and apply $\phi$). Hence $im \ \psi \simeq H$ and so $im \ \psi \times K \simeq H \times K$. Therefore we can use the aforementioned result with $L = im \psi \subset G$ and $K = \ker \phi \subset G$, because $G$ is abelian and so every subgroup is normal. We will have then proved that $G \simeq im \psi \times K \simeq H \times K$, as desired. In effect,

  • let $x \in K \cap L$. Then $x = \psi(h)$ for some $h \in H$ and also $$ 1 = \phi(x) = \phi\psi(h) = id(h) = h $$ which proves that $x = \psi(h) = \psi(1) = 1$ and thus $K \cap L = 1$.

  • take $g \in G$. Now $g = g(\psi\phi(g))^{-1}\psi\phi(g)$. Since $$ \phi(g(\psi\phi(g))^{-1}) = \phi(g)\phi(\psi\phi(g))^{-1}) =\\ \phi(g)\phi(\psi\phi(g)))^{-1} = \phi(g)(\phi\psi\phi)(g)^{-1} = \phi(g)\phi(g)^{-1} = 1, $$ as $\phi\psi\phi = 1_H\phi = \phi$, then $g \in KL$, which concludes the proof.

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