$T$: A set of natural numbers.
$C_1: 2$ is the only prime number that divides elements of $T$.
$C_2 :$ If $i, j \in T$, and $i < j$, then $i$ divides $j$.
For $C_1 \rightarrow C_2$, I think it's true because it's impossible to find a false $C_2$ when $C_1$ is true, i.e., in order for $C_1$ to be true, elements of $T$ must be equal to $2^n$, where $n$ is a natural number. So any $i$ or $j$ where $i$ is less than $j$ in a set of $T$ where elements of $T$ must be equal to $2^n$ can never be false. Is this true?
I'm stuck with $C_2 \rightarrow C_1$ though. Can you please help? Cheers.
Your analysis of $C_1 \to C_2$ is correct.
As Peter remarked in the comments, we can take $T = \{3^n: n \in \Bbb N\}$ to disprove $C_2 \to C_1$.
In fact, by the nature of $C_2$, we can also take $T = \{n\}$ for any $n \ne 2^m$. For then there are no $i, j$ with $i < j$, making $C_2$ vacuously true, while $C_1$ is false.