The following is a lemma in Algebra (page 44): $A$ is a finite abelian p-group. Let $\overline{b}$ be an element of $A/A_1$ ($A_1$ is a cyclic group generated by $a_1 \in A$ of period $p^{r_1}$), of period $p^r$. Then there exists a representative a of $\overline{b}$ in $A$ which also has period $p^r$.

Proof. Let b be any representative of $\overline{b}$ in A. Then $p^rb$ lies in A, say $p^rb = na$ with some integer $n \ge 0$. We note that the period of $\overline{b}$ is $\le$ the period of b. If $n = 0$ we are done.Otherwise write $n = p^k\mu$, where $\mu$ is prime to $p$. Then $\mu a_1$ is also a generator of $A_1$, and hence has period $p^{r_1}$. We may assume $k \le r_1$. Then $p^k\mu a_1$has period $p^{r_1-k}$. By our previous remarks, the element $b$ has period $$p^{r+r_1-k}$$ whence by hypothesis,$\underline{r + r_1 - k \le r_1}$ and $r \le k$. This proves that there exists an element $c\in A_1$ such that $p^rb = p^rc$. Let $a = b - c$. Then $a$ is a representative for $\overline{b}$ in $A$ and $p^ra = 0$. Since period $(a) < p^r$ we conclude that $a$ has period equal to $p^r$ .

My question is, how the underline part in the proof makes sense.

  • 1
    $\begingroup$ What is $\;A\;$, though? $\endgroup$ – DonAntonio Dec 11 '18 at 10:34
  • $\begingroup$ @DonAntonio I have edited $\endgroup$ – Yukinoshita is My Waifu Dec 11 '18 at 10:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.