I am wondering if the following is true:
Let $G$ be a finite group, $g \in G$, and let $\phi : G \rightarrow H$ be a group homomorphism to another finite group $H$. Then $|g| = |\phi(g)| \cdot |k|$ for some element $k$ in $\ker \phi$.
To be specific, the element $k \in \ker \phi$ is taken to be $g^{|\phi(g)|}$. This clearly satisfies $\phi(g^{|\phi(g)|}) = \phi(g)^{|\phi(g)|} = e$.
I came up with a possible proof of this statement, but I wanted to see if anyone had counter-examples or insights on this issue.
The proof I had is to show that $|g|$ divides $|\phi(g) \cdot |k|$ and conversely, that $|\phi(g)|$ divides $\frac{|g|}{|k|}$, using the fact $|x|$ divides $n$ iff $x^n = e$.