Dtjytyk

Looking for a validation of a proof of a stronger statement. The essential argument that needs review is the final paragraph. This would obviate the need for a much more complicated argument I gave at Correcting Proof of Lemma 10 in Kaplansky's *Infinite Abelian Groups* as this version of Lemma 7 would suffice in the proof of Lemma 10.

The lemma states:

Let $G$ be a primary group and $S$ a subgroup with no elements of infinite height. Suppose that the elements of order $p$ in $S$ have the same height in $S$ as in $G$. Then $S$ is pure.

Here "primary group" means "abelian p-group". The height of $x$ in $G$ (denoted $h_G(x)$) is the supremum of all $n$ such that there exists a $y$ with $x=p^ny$. $S$ is pure if the height of every element of $S$ is the same in $S$ as it is in $G$.

I claim that the hypothesis that $S$ have no elements of infinite height is unnecessary. We can state:

Let $G$ be a primary group and $S$ a subgroup. Suppose that the elements of order $p$ in $S$ have the same height in $S$ as in $G$. Then $S$ is pure.

Proof (following Kaplansky quite closely, but adding (II)): Induct on the order of $x$. If it is true that all elements of $S$ of order $p^n$ or less have the same height in $S$ and $G$, let $x$ have order $p^{n+1}$. Then $px$ has the same height in $S$ and $G$; denote that common height by $r$, which may be $infty$, and is at least $1$.

(I) If $r<infty$ then $px=p^ry$ for some $yin S$. Then the height of $p^{r-1}y$ must be $r-1$ even in $G$, lest $px$ have height more than $r$. Note that $x=(x-p^{r-1}y)+p^{r-1}y$. Since $p(x-p^{r-1}y)=0$, $x-p^{r-1}y$ has the same height $k$ in $G$ and in $S$. If $kne r-1$, even if $k=infty$, then $h(x)=min(k,r-1)$ in both $G$ and $S$. If $k=r-1$, then $h(x)=r$ in both $S$ and $G$, since it must be at least $r$ as the sum of elements of height $r$, and it must be at most $r$ lest $px$ have height greater than $r$.

(II) If $r=infty$ then for any positive integer $m$ there exists a $yin S$ with $px=p^my$. The height of $p^{m-1}y$ must be at least $m-1$ in both $G$ and $S$, though it need not be the same in both. The height of $x-p^{m-1}y$, call it $j$, must be the same in $G$ and $S$ since it has order $p$. Thus, if $j<m-1$, the height of $x$ must be exactly $j$ in both $G$ and $S$. And if not, the height of $x$ is at least $m-1$ in $S$. But if $x$ has finite height $i$ in $S$, we can choose $m>i+1$, and we see that $x$ also has height $i$ in $G$. On the other hand, if $x$ has infinite height in $S$ it clearly also has infinite height in $G$.