# Show $K\times H\cong G$

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.

• 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 '18 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 '18 at 14:20
• 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 '18 at 14:46
• That's an awesome statement! I'm wondering where can we use this result? – nafhgood Dec 6 '18 at 15:05
• @mathnoob This result is called the splitting lemma and since it has a name, it's important. – Arthur Dec 6 '18 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.