# Generalized height of elements in abelian groups

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.

• I think you meant to write $0=h^*_p(f(a))\leq h^*_p(a)$? – Jeremy Rickard Jun 27 '16 at 10:43
• The statement is "$h^*_p$ does not diminish under homomorphisms" so $h_p^*(f(a))\geq h_p^*(a)$ – Ilan Barnea Jun 27 '16 at 10:46
• Oh, OK. You meant that's the inequality that would be true if the statement were true, rather than the inequality that is true. – Jeremy Rickard Jun 27 '16 at 10:48
• Exactly, this shows that the statement cannot be true – Ilan Barnea Jun 27 '16 at 10:50
• I've not checked that this fixes all the problems, hence just a comment. But the idea of the definition of $h^*_p$ seems to be that if there is no $\sigma$ with $a\in p^\sigma A\setminus p^{\sigma+1} A$ then $h^*_p(a)$ should be "big", and $l_p(A)$ is chosen as the smallest ordinal bigger than $h^*_p(b)$ for all those $b\in A$ which already have $h^*_p(b)$ defined. He'd probably have chosen $\infty$ instead of $l_p(A)$ except that then you have to say which infinite ordinals are greater than $\infty$. But if you do choose $\infty$, a notional symbol greater than all ordinals, does that work? – Jeremy Rickard Jun 27 '16 at 13:54

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 $\infty$).
• Is the existence of $\infty$ compatible with axioms from set theory ? For example why can't one define $\infty + 1$ ? – tj_ Jun 28 '16 at 14:19
• @tj_ It's nothing to do with the axioms of set theory. All I'm saying is that there is one case in which the generalized $p$-height is not an ordinal, and in that case we call the $p$-height $\infty$ and deem that to be greater than all ordinals. If you prefer, we could just say that it's not defined, and make the necessary but annoying adjustments to all statements about the ordering of generalized $p$-heights. – Jeremy Rickard Jun 28 '16 at 17:29