Consider the twice punctured sphere. Now I know the mapping class group of the twice punctured sphere is $\mathbb{Z}_{2}$, the cyclic group of order 2. However, I know one applies the alexander trick to show that something is the identity. Is the idea as follows :
You can think of the punctures as $\mathbb{Z}_{2}$ since we can either just leave the punctures as is or switch them. The question is, what happens with everything else. i.e we know that the mapping class group is $\mathbb{Z}_{2} \times$ something, and if we can just show that $e$ the identity is that something then it will follow that the mapping class group of the twice punctured sphere is just $\mathbb{Z}_{2} \times e = \mathbb{Z}_{2}$.
Is this the right idea? How can I be more specific when I say "something".
If $G$ is the mapping class group, what you've (almost) proven is that you a surjective morphism $\phi: G \to \mathbb{Z}_2$ where you map $f$ to $1$ if it switches the punctures and $0$ if it leaves them fixed. To prove that completely, you need to show (it's very easy) that this is compatible with composition and isotopy.
Once you've done that, you know that $G / \operatorname{ker}(\phi) \cong \operatorname{im}(\phi) = \mathbb{Z}_2$. So if you prove that $\operatorname{ker}(\phi)$ is trivial, in other words that any automorphism that fixes the punctures is isotopic to the identity, then you're done. This is where the Alexander trick comes into play.
In general you can't say outright that $G = \mathbb{Z}_2 \times H$ for some $H$ though. It could be for example that $G = \mathbb{Z}$ and $\phi$ is reduction mod 2. If you want to get fancy you can say that you have an exact sequence $$ 1 \to H \to G \to \mathbb{Z}_2 \to 1$$ where $H$ is some group (the kernel of $\phi$). It's not always the case that $G$ is then a product.