Fundamental Theorem of Abelian Groups - intuition regarding Lemma

Question

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.

Answer 1

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

Answer 2

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 $mc=0$.