Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $G$ be a finite abelian $p$-group. It is quite elementary to see that if $g \in G$ is an element of maximal order (and thus its span is a cyclic subgroup of $G$ of maximal order) then $G$ can be written as the direct sum $G=\langle g \rangle \oplus H$ for some $H \leq G$ (subgroup of $G$). For a proof see this for example (page 2).

My question: Do we need that $G$ is a $p$-group or does it also work for arbitrary finite abelian groups?

I think it is wrong for general groups because I looked around quite a bit and always only found the above theorem, but I could not find a counter-example.

share|cite|improve this question
up vote 8 down vote accepted

The result follows for arbitrary finite abelian groups from the $p$-group case.

Remember that a finite abelian group $G$ is the direct sum of its $p$-parts, $$G = G(p_1)\oplus \cdots\oplus G(p_n),$$ where $p_1,\ldots,p_n$ are the distinct primes that divide $|G|$, and $$G(q) = \{ a\in G\mid q^ma = 0 \text{ for some }m\geq 0\},\qquad q\text{ a prime.}$$

If $a\in G$ is of maximal order, then we can write $a=a_1+a_2+\cdots+a_n$, where $a_i\in G(p_1)$. Since $a$ is of maximal order in $G$, then $a_i$ is of maximal order in $G(p_i)$. By the $p$-group case, we can write $G(p_i) = \langle a_i\rangle\oplus H_i$ with $H_i\leq G(p_i)$. Then $H_1+\cdots+H_n$ is a subgroup of $G$, it is the internal direct sum of the $H_i$, and since $G(p_i) =\langle a_i\rangle\oplus H_i$, then $$\begin{align*} G &= G(p_1)\oplus \cdots \oplus G(p_n)\\ &= (\langle a_1\rangle\oplus H_1) \oplus \cdots \oplus (\langle a_n\rangle \oplus H_n)\\ &= (\langle a_1\rangle\oplus\cdots \oplus\langle a_n\rangle) \oplus (H_1\oplus\cdots\oplus H_n). \end{align*}$$ To finish off, note that $\langle a_1\rangle\oplus\cdots\oplus \langle a_n\rangle = \langle a\rangle$ (e.g., by the Chinese Remainder Theorem).

share|cite|improve this answer
Great to have this answered by someone who actually researches on p-groups, thank you a lot :-) – Listing Nov 25 '11 at 21:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.