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.

wouldbe true if the statement were true, rather than the inequality thatistrue. $\endgroup$ – Jeremy Rickard Jun 27 '16 at 10:48