# Question in Algebra by Serge Lang.

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.

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