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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,

Kasp9201.

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    $\begingroup$ 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
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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 $G= \prod_{p^k \mid\mid n} G(p^k)$.

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  • $\begingroup$ 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
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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 $mc=0$.

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  • $\begingroup$ 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

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