Why $(\mathbb{Z}/N\mathbb{Z})^{\times}[2]$ is of order $2$?

Question

Why if $(\mathbb{Z}/N\mathbb{Z})^{\times}$ is cyclic, the group of his elements of order dividing $2$ is of order 2?

Answer 1

You need also $N>2$. In a cyclic group, for each divisor $d$ of the order of the group there are exactly $\varphi(d)$ elements of order $d$. Since $\varphi(N)$ is even for all $N>2$ your group $(\mathbb{Z}/N\mathbb{Z})^\times$ has even order so that $2$ is a divisor. Then you have elements of order $1$ and $2$ to count (these are the ones whose order divides 2). We have only the identity of order 1, and $\varphi(2)=1$ element of order 2, so exactly 2 elements.

Answer 2

Take the cyclic group $x^0,x^1,x^2$ the only element with order dividing $2$ is is $x^0$ since $(x^1)^2=x^2$ and $(x^2)^2=x^1$. So the subgroup of the cyclic group of order $3$ that consitsts of the elements which have order dividing $2$ is not of order $2$.

In fact your claim is only true for cyclic groups of even order.

However since the order of $(\mathbb{Z}/N\mathbb{Z})^{\times}$ is even for $N\neq 1,2$ the claim is true.