Let $G$ be an abelian $p$-group, meaning that every element has order $p^n$ for some $n$. We can define, inductively, $G_0 = G$, $G_{\alpha + 1} = pG_{\alpha}$, and $G_{\alpha}$ for limit $\alpha$ is defined in the obvious way by taking intersections. For $x \in G$, we define the height of $x$, denoted $h(x)$, to be $\infty$ if $x=0$ and $\alpha$ if $x \in G_{\alpha}$ but not $G_{\alpha + 1}$. It can be easily shown that if $h(x) \neq h(y)$ then $h(x+y) =$ min $(h(x),h(y))$, if $h(x) = h(y)$ then $h(x+y) \geq h(x)$, and if $x \neq 0$ then $h(px) > h(x)$.
Is it true that $h(px) = h(x) + 1$?
If you happen to be interested in why I am asking, it is because I saw this claim made in the book "A course in the theory of Groups (Second Edition)" by Derek Robinson (page 115 to be exact.) However, it might be problematic for other reasons. (Edited since the exact reason I originally had in mind isn't actually a problem.)
It's not true that $h(px)=h(x)+1$ in general, although of course $h(px)\geq h(x)+1$. For example, take $G=C_{p^n}\times C_p$, generated by $(x,y)$, and consider $g=x^{p^{n-2}}y$. Then this has height $0$ because there's no $p$th root of $y$. But $g^p=x^{p^{n-1}}$, which has height $n-1$ (as $x^{p^{n-1}}$ has order $p$ in a group of order $p^n$).