Showing that the extension $1 \to C_3 \to C_6 \to C_2 \to 1$ is split by finding complement of $C_3$ I'm trying to solve the following exercise:

Show that the extension $1 \to C_3 \to C_6 \to C_2 \to 1$ is split but the extension $1 \to C_2 \to C_4 \to C_2 \to 1$ is not split. 

I think one way to show that an exact sequence $1 \to N \to G \to Q \to 1$ is split is to construct a valid traversal function $\tau: Q \to G$ which is a group homomorphism. The other way is to show that $N$ has a complement in $G$. I'm trying to follow the second method here. 
My attempt
Showing $1 \to C_3 \to C_6 \to C_2 \to 1$ is split
Let $C_2 = \{e, a\}$ where $a$ is a generator of $C_2$.
Let $C_3 = \{e, b, b^2\}$ where $b$ is a generator of $C_3$.
I need to show that $C_3$ has a complement in $C_6$ which is isomorphic to $C_2$ (?). 
Also, we know that the product of two subgroups $S$ and $T$ is a group if and only if $ST = TS$. But I'm stuck here, i.e., when trying to show that $C_2C_3 = C_3C_2$.
$$C_2C_3 = \{e, b, b^2, a, ab, ab^2\}$$
$$C_3C_2 = \{e, a, b, ba, b^2, b^2a\}$$
The elements $e, a, b, b^2$ are common in both the sets. But then, how do I show that $ab=ba$ or $ab = b^2a$?
 A: This doesn’t work as you seem to hope. For one thing, $C_2$ is given as a quotient of $C_6$; you would need to define a map from $C_2$ to $C_6$ in order to talk about the image of $a$ in $C_6$.
But in order to do that you must first define the splitting function, which is what you want to do in the first place. So I don’t see you hope to do what you want to do. Note that there is a group that is a non-split extension of $C_3$ by $C_2$, in which $ab$ is not going to equal $ba$; namely, $S_3$. So what you are trying to do just isn’t going to work.
You have to use the fact that you are working with $C_6$ (and later, with $C_4$).
If $C_6 = \{1,x,x^2,x^3,x^4,x^5\}$, then there are only two possible images for $b$; and there is only one possible image for $a$ if you are going to have a group morphism. You should be able to take it from there.
In contrast, with $C_4=\{1,y,y^2,y^3\}$, there is only one possible splitting for $C_2$, but that does not yield a complement to the only possible image of $C_2$.
A: The two sets $C_2C_3$ and $C_3C_2$ both have $6$ elements, and they are both subsets of $C_6$, which also has $6$ elements. So all three of these sets have to be equal.
