In this question Parity of members in a group I defined even members of a group $G$ as all members $b \in G : b \neq a^ca^{c+1}$ where $a \in G$ and $c \in \mathbb{N}$ . This follows from the fact that all even numbers in the set of integers are those members that are not the sum of two consecutive numbers. Now comparing to an alternative definition of even members of a group; that is the even members of a group $G$ are all members $b \in G : b=a^{2c}$ where $a \in G$ and $c \in \mathbb{N}$.
Consider a cyclic group $<x>$. If the order of $x$ is odd, then the two definitions of even members produce different subsets of $<x>$ which are the even members. However if the order of $x$ is even, the definitions produce the same subsets of $<x>$ as the even members. So the group $(\mathbb{Z}, +)$ must have an even order, $\aleph_0$ is even! Furthermore by similar arguments for trisectable members (divisible by $3 $ and the sum of three consecutive members), we can show that $\aleph_0$ is trisectable, and with further arguments that it is divisible by any prime number!
How true is this?
I think you're doing some invalid generalizations.
Things that are true regarding order of finite cyclic groups for example does not have to hold for an infinite cyclic group. That is for example the observation that your two definitions agree only for groups of even order is true when you have the standard definition of the order of a group, but when you extend that definition it doesn't follow that it is true.
Extending the concept of divisibility seem possible, you can still do the same claim that a cardinality is divisible by another if you could divide the set in that number of equally sized parts. But that definition breaks down when extending it to infinite sets since it will mean that $\aleph_0$ will be divisible by every positive integer and $\aleph_0$ itself. I'd guess that it's true for any $c\ge\aleph_0$ that it is divisible by all $d \le c$.