Find the identity element of $2\mathbb{Z}_6$ under multiplication. The first thought I had was to construct a Cayley table, but then it dawned on me that I have no clue how to construct one. Here is what my intuition is telling me: $a\mathbb{Z}_{ap}$ has an identity element because $\gcd(a,p)=1$ and $p$ is prime. Then, $a\mathbb{Z}_{ap}$ is a group modulo $ap$ where all elements are multiples of $a$. The table looks like:
$$\begin{array}{|c|c|c|c|c|c|c|} 
\mod12 & \text{2} & \text{4} & \text{6}& \text{8}& \text{10}& \text{12}\\ \hline
\text{2} &  &  \\ \hline
\text{4} &  &  \\ \hline
\text{6} &  &  \\ \hline
\text{8} &  &  \\ \hline
\text{10} &  &  \\ \hline
\text{12} &  &  \\ \hline
\end{array}$$
I feel like this is majorly flawed, because attempts at filling in the table yields to all types of issues and what not. Any help is greatly appreciated!
Follow-up question: Is there a way to find the identity element of $a\mathbb{Z}_{ap}$ without needing to construct a Cayley table?

The title question:

Find the identity element of $2\mathbb{Z}_6$ under multiplication.

The set you are talking about is $\{0, 2, 4\}$, where each of these numbers actually stands for a coset of $\mathbb Z$ modulo $6\mathbb Z$.
Since $2\cdot 4=8\equiv 2$ and $4\cdot 4=16\equiv 4$, $4$ is obviously an identity for this subring. 
