Definitions: A continuous dynamical system $(W^t, X)$ is defined with the function $W^t:X \rightarrow X$ satisfying $W^t \circ W^s = W^{t+s}$. A point $x \in X$ is periodic if $W^p(x) = x$ for some $p \ne 0$ and the minimum $p$ satisfies this is the period of $x$.
Problem: I want to show that if the time-1 map (i.e. the discrete dynamical system $(T, X)$ with $T = W^1$) of $W^t$ has a periodic point $x_0$ with period $p \ge 2$ (i.e. $p$ is the minimum integer with $T^p(x) = x$) then it must have infinite number of periodic points in $X$ with the same period.
I can show that all of the points $x_r = W^r(x_0)$ with $r < p$ has period $p$. However now I need to show that there are infinitely many $W^r(x_0)$s, i.e the set $\{W^r(x_0) | r \in \mathbb{R}\}$ is infinite. My current approach is to show that $W^n(x_0) \ne W^m(x_0)$ for all real number $n \ne m$ and $n, m < p$, which I was unable to do. I wasn't even able to show the discrete case since the definition of period only granted one $y = T^k(x_0) \ne x_0$ with $k < p$, but could it be the case that $T^k(x_0) = y$ for all $k < p$?
So I am not sure whether my current approach is even correct or not. Help is appreciated!