Two dynamical systems perpendicular I am given two dynamical systems:
$$x'=f(x)$$
$$x'=g(x)$$
where $f:\mathbb{R}^2\rightarrow \mathbb{R}^2$ and $g:\mathbb{R}^2\rightarrow \mathbb{R}^2$ are $C^1$ and perpendicular ($\langle f(x),g(x)\rangle=0$ for all $x's$). I am to show that if one of the systems has a (nontrivial) periodic orbit, then the other has a fixed point. Any help with this problem would be appreciated.
Suppose without loss of generality hat $f$ has a nontrivial periodic orbit. Consider the path $\gamma$ that describes the orbit. $\gamma$ is closed because the orbit is periodic; moreover, by the Existence and Uniqueness Theorem for ODEs, $\gamma$ must be simple. It is hence a Jordan curve and we may speak of the interior and exterior of $\gamma$.


  EDIT: If $g$ vanishes at some $x\in\gamma$, then $x$ is a trivial fixed point of the system induced by $g$. Assume then that $g$ is nowhere vanishing on $\gamma$.


Because $g$ is continuous, always orthogonal to $f$ and nowhere vanishing on $\gamma$, along $\gamma$ the field $g$ must face either always inwards (towards the interior) or always outwards (towards the exterior).

If $g$ faces always inwards, consider what happens as the time goes to $+\infty$. If $g$ faces always outwards, consider what happens as the time goes to $-\infty$. Do you think you can take it from here?

Hint: Choose a point $p\in\gamma$ and consider its orbit under $g$. Can the orbit intersect $\gamma$ at some point other than $p$? Use the existence and uniqueness theorem!
Suppose without loss of generality hat $f$ has a nontrivial periodic orbit. Consider the path $\gamma$ that describes the orbit. $\gamma$ is closed because the orbit is periodic; moreover, by the Existence and Uniqueness Theorem for ODEs, $\gamma$ must be simple. It is hence a Jordan curve and we may speak of the interior and exterior of $\gamma$.


  EDIT: If $g$ vanishes at some $x\in\gamma$, then $x$ is a trivial fixed point of the system induced by $g$. Assume then that $g$ is nowhere vanishing on $\gamma$.


Because $g$ is continuous, always orthogonal to $f$ and nowhere vanishing on $\gamma$, along $\gamma$ the field $g$ must face either always inwards (towards the interior) or always outwards (towards the exterior).

If $g$ faces always inwards, consider what happens as the time goes to $+\infty$. If $g$ faces always outwards, consider what happens as the time goes to $-\infty$. Do you think you can take it from here?

Hint: Choose a point $p\in\gamma$ and consider its orbit under $g$. Can the orbit intersect $\gamma$ at some point other than $p$? Use the existence and uniqueness theorem!