In the book *Infinite abelian groups Vol. I* by L. Fuchs, on page 154, the notion of the generalized $p$-height of an element in an abelian group is defined, as follows: 


Let $A$ be an abelian group and let $p$ be a prime number. First we define for every ordinal $\sigma$ a subgroup $p^\sigma A$ of $A$, recursively, by letting $p^0 A:=A$, $p^{\sigma+1}A:=p(p^\sigma A)$ for every ordinal $\sigma$ and 
$$p^\sigma A:=\bigcap\_{\lambda<\sigma}p^\lambda A$$
for every limit ordinal $\sigma$. The smallest ordinal $\sigma$ for which $p^{\sigma+1}A=p^\sigma A$ is called the $p$-length of $A$ and denoted $l\_p(A)$.


Now let $a$ is an element of $A$. If $a\in p^{l\_p(A)}A$, we define $h^\*\_p(a):=l\_p(A)$. Otherwise, there exists a unique ordinal $\sigma$ for which $a\in p^\sigma A\setminus p^{\sigma+1} A$, and we define $h^\*\_p(a):=\sigma$. $h^\*\_p(a)$ is called the *generalized $p$-height* of $a$.


After giving this definition it is claimed (see (v) on page 154) that $h^\*\_p$ does not diminish under homomorphisms. Clearly this statement is not true as stated since for any abelian group $A$ we can consider the homomorphism $f:A\to 0$, so for every $a\in A$ we would obtain $0=h^\*\_p(f(a))\geq h^\*\_p(a)$.


My question is what would be a correct reformulation of this statement? Please also indicate the proof. I am asking because this statement is used in the proof of Lemma 37.1 that follows. 



Generalized height of elements in abelian groups
As suggested in my comment, define $h^\*\_p(a)$, as Fuchs does, to be the smallest ordinal $\sigma$ with $a\not\in p^{\sigma+1}A$ if there is such a $\sigma$, but if there is no such $\sigma$ then define $h^\*\_p(a)=\infty$, where $\infty$ is just a symbol that is defined to be greater than every ordinal.


If $\varphi:A\to B$ is a homomorphism, then since $\varphi(pA')\leq p\varphi(A')$ and $\varphi\left(\bigcap\_iA\_i\right)\leq\bigcap\_i\varphi(A\_i)$ for subgroups $A',A\_i$ of $A$, a straightforward transfinite induction shows that $\varphi(p^\alpha A)\leq p^\alpha B$ for every ordinal $\alpha$, and so $h^\*\_p\left(\varphi(a)\right)\geq h^\*\_p(a)$ for every $a\in A$ (including the possibility that $h^\*\_p(a)<\infty$ but $h^\*\_p\left(\varphi(a)\right)=\infty$, or that $h^\*\_p(a)=\infty$, in which case $h^\*\_p\left(\varphi(a)\right)$ is also necessarily $\infty$).


