I’m reading about group extensions and I’m finding hard to understand how the following definition (from Wikipedia):
A split extension is an extension $1\to K\to G\to H\to 1$ with a homomorphism $s\colon H\to G$ such that going from $H$ to $G$ by $s$ and then back to $H$ by the quotient map of the short exact sequence induces the identity map on $H$ i.e., $\pi \circ s={\mathrm {id}}_{H}$. In this situation, it is usually said that $s$ splits the above exact sequence.
is related to this one (from my lecture notes):
An extension is called split if $G$ has a subgroup isomorphic to $H$ which intersects trivially with $K$.
Is there any easy way to see the intuition behind both definitions?
I’ve tried applying them to understanding why the quaternion group $H_8$ is a non-example: you can give it the extension structure $1 \to C_4 \to H_8 \to C_2 \to 1$, but I’m being told this extension isn’t split. Using the second definition, I can see that the subgroup ${-1,1}$ is isomorphic to $C_2$ and its intersection with the subgroups $\langle i\rangle $, $\langle j\rangle $ and $\langle k\rangle $ (isomorphic to $C_4$) is never trivial (because the element $-1$ is in all of them), so that makes sense. But I don’t know how to give a similar explanation using the first definition.
Given any extension, $$1\to K\to G\stackrel \pi\to H\to 1,$$ you have in some sense broken down the group $G$ into the groups $K$ and $H$. This is a bit like factoring numbers.
Whilst for any such extension we may identify $G$ and $K\times H$ as sets (assuming the axiom of choice), if the extension splits, the splitting gives us a explicit identification of the sets $G$ and $K\times H$. Whilst this identification does not preserve the group structure, it comes close.
Let's start with the first definition, and then we will see what the problem with the second definition is.
So we have a homomorphism $s\colon H \to G$, with $\pi \circ s=1_H$. As $s$ has a left inverse, it must be injective. We therefore have the subgroup $s(H)\subset G$, isomorphic to $H$.
We denote all identity elements of groups by $e$. The intersection $s(H)\cap K$ is just $\{e\}$, as if $x\in s(H)\cap K$, then $x=s(h)$ and $\pi(x)=e$, so $$h=\pi s(h)=\pi(x)=e,$$ and $x=s(h)=s(e)=e$.
So we have these two groups $K,s(H)$ sitting inside $G$ and intersecting trivially. Now take any $g\in G$. We have $h=s\pi(g)\in s(H)$ and $$\pi(h)=\pi(s\pi(g))=\pi(g).$$ Let $k=gh^{-1}$, so $g=kh$. Then $$\pi(k)=\pi(gh^{-1})=\pi(g)\pi(g)^{-1}=e,$$ and we may conclude that $k\in K$. Thus any $g\in G$ satisfies $g=kh$, with $k\in K, h\in s(H)$. Further this factorisation is unique: \begin{eqnarray*}k_1h_1=k_2h_2 &\implies& k_2^{-1}k_1=h_2h_1^{-1}\in K\cap s(H)=\{e\}\\&\implies& k_1=k_2,\,\,h_1=h_2.\end{eqnarray*}
So we have identified $G$ and $K\times H$ as sets. However they are not necessarily the same as groups. In $G$ we have the following product: $$ (k_1h_1) (k_2h_2)=k_1(h_1k_2h_1^{-1}) h_1h_2.$$
So we have established that the fist definition gives us a subgroup of $G$, isomorphic to $H$ and intersecting trivially with $K$. So what can go wrong with the second definition?
The issue here, is that given a subgroup $H'\subseteq G$ isomorphic to $H$ and intersecting $K$ trivially, we may have $\pi(H')$ a proper subgroup of $H$. For example the even numbers are a proper subgroup of $\mathbb{Z}$, isomorphic to $\mathbb{Z}$. So for the definition to be correct, we cannot just have any old subgroup $H'$ isomorphic to $H$ and intersecting $K$ trivially, but one where $\pi$ restricts to an isomorphism $H'\to H$.
In your example, let $i$ generate $C_4$. Then under any splitting $s$, the generator of $C_2$ must map to something that maps back to it, so it must go to $\pm j$ or $\pm k$. However, in order for $s$ to be a group homomorphism, it must take elements of order two, to elements of order two. None of $\pm j$ or $\pm k$ have order two - they all have order four. So there is no splitting homomorphism $C_2\to H_8$ in this case.
This is the general situation - we always have a map of sets $H\to G$ which is a right inverse to $\pi$. However to have a splitting, we must be able to choose such a map which is also a group homomorphism.
An example of an extension which satisfies the second definition but is not split:
$$ 1\to \left(\prod_{i=1}^{\infty} C_2\right)\times\left(\prod_{i=1}^{\infty} C_2\right)\stackrel\iota\to \left(\prod_{i=1}^{\infty} C_2\times C_2\right)\times \left(\prod_{i=1}^{\infty} C_4\right)\stackrel\pi\to \left(\prod_{i=1}^{\infty} C_2\right)\times \left(\prod_{i=1}^{\infty} C_2\right)\to 1 $$
Here $\iota$ is the product of inclusions $C_2\to C_2\times C_2$ and inclusions $C_2\to C_4$.
Then in $\left(\prod_{i=1}^{\infty} C_2\times C_2\right)$ we have a copy of $\left(\prod_{i=1}^{\infty} C_2\right)$ which intersects the image of $\iota$ trivially. This subgroup is isomorphic to $\left(\prod_{i=1}^{\infty} C_2\right)\times \left(\prod_{i=1}^{\infty} C_2\right)$. Thus the second definition is satisfied.
However this sequence is not split. If $s$ were a splitting, and $t$ a generator of a copy of $C_2\cong C_4/C_2$ on the right, then $s(t)$ should be an element in the middle group which has order two, and maps to $t$. However the only elements which map to $t$ are of order four.