I'm learning group theory and have come across the Fundamental Theorem of Abelian Groups. My material is a book by Thomas W. hungerford called "Abstract Algebra: An Introduction". In this, Hungerford proves the theorem using a variety of lemmas. One of the lemmas (Lemma 9.3 in my book) goes like this:

"Let $G$ be an abelian group an $a \in G$ an element of finite order. Then $a = a_1 + a_2 + ... + a_t$ with $a_i \in G(p_i)$, where $p_1,p_2,...,p_t$ are the distinct positive primes that divide the order of $a$."

In which we have defined $G(p_i)= \lbrace a \in G : \vert a\vert = np_i, n \in \mathbb{Z} \rbrace$

Hungerford proves this lemma by induction. I understand the proof, but it has not made it clear to me on an intuitive level why we may express $a \in G$ in such a way. I understand my question is vague, but it is because I'm not sure what exactly it is that I'm looking for. Thus, I would like to ask if you smart people would explain your intuitive understandings of the lemma. Then maybe I will see the light myself.

Thank you very much,


  • 1
    $\begingroup$ I know it is basically cheating, but the "intuition" part is what leads you, in the end, to the Fundamental Theorem itself: in the end, that the group can be expressed as a product of cyclic groups, separating each prime from the other (since $C_{pq} = C_p \times C_q$ if $(p,q)=1$). This lemma is "technical" but its intuitive meaning is what you will discover with the theorem. $\endgroup$ – AnalysisStudent0414 Mar 12 '18 at 14:53
  • $\begingroup$ Ah, yes. If I understand your reply correctly, then this is exactly what I have so far, but also what I am dissatisfied with. However, you made me realize what it is that I should actually be asking: What intuition leads us to this lemma? And thank you very much for your reply $\endgroup$ – kasp9201 Mar 12 '18 at 15:04

Here is some intuition.

Let $G$ be an abelian group of order $n$. If we can write $n=rs$ with $\gcd(r,s)=1$, then $G = G(r) \times G(s)$, where $G(m) = \{ g \in G : g^m =1 \}$.

Indeed, write $1 = ru + sv$. Then $g = g^1 = g^{ru + sv} = g^{ru}g^{sv}$ and $g^{ru} \in G(s)$ and $g^{sv} \in G(r)$. Finally, $G(r) \cap G(s)=1$, again because $\gcd(r,s)=1$.

Therefore, you can decompose $G= \prod_{p^k \mid\mid n} G(p^k)$.

  • $\begingroup$ I hate it when this happens... just for the record, I was typing my answer when yours appeared. $\endgroup$ – David C. Ullrich Mar 12 '18 at 15:28
  • $\begingroup$ @DavidC.Ullrich it's good to have both explanations. $\endgroup$ – lhf Mar 12 '18 at 15:29
  • $\begingroup$ Great, I've got two nice explanations now. Unfortunately I have to choose, so I will accept yours as the answer to my question, as it was the first to be posted. So thank you very much, I understand the lemma better now with the help of David and you $\endgroup$ – kasp9201 Mar 12 '18 at 15:35

I don't know if this will seem more "intuitive" or not, but it's a proof that's not by contradiction.

What seems intuitive depends on what you know. I've been teaching myself a little algebra, and I was delighted to discover, on reading your question, that I actually knew a proof. It's based on the following:

If $n,m$ are relatively prime integers then there exist integers $j$ and $k$ with $jn+km=1$.

If you can convince yourself that that is "intuitively" true then you're set. (How to make that seem "intuitive": Read the proof, study the proof, contemplate the proof til it seem obvious.)

Now about abelian groups: First note you mix additive and multiplicative notation in your question. I'm going to stick to additive notation; hence where you write $a^n$ I'll write $na$.

The result you ask about is immediate from the following, by induction on the number of prime factors of the order:

Suppose $n,m$ are relatively prime integers, $a$ is an element of an abelian group, and $nma=0$. Then $a=b+c$ where $nb=0$ and $mc=0$.

Proof: Choose $j,k$ so $jn+km=1$. Then $a=b+c$ if $b=kma$ and $c=jna$. And $nb=k(nma)=0$; similarly $mc=0$.

  • $\begingroup$ Ah, neat. Thank you for your heads up on mixing with additive and multiplicative operations. I am, however, going to choose the answer of lhf as he was right before you. Nevertheless, your answer also added to my understanding, so thank you very much! $\endgroup$ – kasp9201 Mar 12 '18 at 15:30

Your Answer

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

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