Statement: Let G be an Abelian group of prime-power order and let a be an element of maximum order in G. Then G can be written in the form $a \times K$.
Proof: We denote |G| by p^n and induct on n. If n = 1, then G = $\langle a \rangle \times \langle e \rangle$. Now assume that the statement is true for all Abelian groups of order $p^k$, where $k < n$.
Among all elements of G, choose $a$ of maximum order $p^m$. [Question: Why are we picking any random m for the order of a? For Abelian groups, isn't the converse of Lagrange always true? So shouldn't the element a have order $p^{k-1}$?]
Then $x^{p^m} = e$ for all $x$ in G. [Question: How, if |G| = p^k?]
We assume that $G \neq \langle a \rangle$, as there would be nothing left to prove then.
Now among all elements of G, choose $b$ of smallest order such does $b \not\in \langle a \rangle$.
We claim that $\langle b \rangle \cap \langle a \rangle = e$
Since $|b^p| = |b|/p$, we know that $b^p \in \langle a \rangle$ by the manner in which b is chosen. [Question: What does he mean by the manner in which b is chosen? Why does b^p lie in the set generated by a? Is it because the group with the minimum order has to have order p according to Lagrange and this straightaway implies b^p = e?]
Say $b^p = a^i$. Notice that $e=b^{p^m}=(b^p)^{p^{m-1}} = (a^i)^{p^{m-1}}$, so $|a^i| \leq p^{m-1}$.
Thus, $a^i$ is not a generator of . Therefore, $gcd(p^m,i) \neq 1$. This proves that $p$ divides $i$. Let $i = pj$. Then $b^p = a^i = a^{pj}$.
Consider the element $c = a^{-j}b$. $c$ is not in .
$c^p = a^{-jp}b^p$. Thus $c$ is an element of order p such that c is not in . Since b was chosen to have minimal order, we can finally say that b has an order of p.
It now follows that $\langle a \rangle \cap \langle b \rangle = {e}$, because any non-identity element of the intersection would generate $\langle b \rangle$ and thus contradict that b lies in .
Consider the factor group $\overline{G} = G/\langle b \rangle$. Let $\overline{x}=x\langle b \rangle$ in G.
If $|\overline{a}| < |a| = p^m$, then $\overline{a}^{p^{m-1}} = \overline{e}$. [Question: here we have picked $p^{m-1}$ just as any arbitrary number less than $p^m$, but still a power of $p$, right?]
This means that $(a\langle b \rangle)^{p^{m-1}} = a^{p^{m-1}}\langle b \rangle = \langle b \rangle$. This implies that the order of a is $p^{m-1}$, which is absurd. So order of a-bar is equal to order of a which is $p^m$.
Therefore $\overline{a}$ is an element of maximum order in $\overline{G}$.
By induction, we know that $\overline{G} = \langle a \rangle \times \overline{K}$. [Question: ??? What induction? What is the basis of this step? What even is K? How are we even defining K in the steps below?]
Let K be the pullback of $\overline{K}$ under the natural homomorphism from G to G-bar.
We claim that $\langle a \rangle \cap K = {e}$. For if $x \in \langle \overline{a} \rangle \cap \overline{K} = {e} = \langle b \rangle$ and $x \in \langle a \rangle \cap \langle b \rangle = {e}$.
It now follows from an order argument (???) that $G = <a>K$, and therefore $G = \langle a \rangle \times K$.
I am currently teaching myself abstract algebra and real analysis and this proof has be confused for a while now. I apologize for the length of the post, but I could not think of any other way to convey my doubts in any concise manner.
Best Answer
This is Lemma 2 Ch. 11 from : 'Gallian , Contemporary abstract algebra' leading to the fundamental theorem.
The induction hypothesis assumes that the lemma has already been proven for all groups of order $p^k $. By contrast $p^m$ is the highest order of all the elements of $G$. So the only thing we can say about $m$ is that it must be smaller than or equal to $n$.
$|G|=p^n$ , not $p^k$ . The orders of all elements are $p^j$ for some $j$ depending on the element. Maximum of these $j$'s is $m$ , by definition. We can write : $x^{p^m}= (x^{p^j})^{p^{m-j}} = e^{p^{m-j}} =e$
Let the order of $b$ be $p^j$. This implies : $\frac{|b|}{p}=p^{j-1}$ , also $b^{p^j}=e \implies (b^p)^{p^{j-1}} =e $. If there where $i<j$ for which $(b^p)^{p^{i-1}} =e $ that would contradict the fact that $j$ is the smallest integer for which $b^{p^j}=e$. So $\frac{|b|}{p}=|b^p|$. We chose $b$ to be the element of smallest order not in $\langle a \rangle$. The order of $b^p$ is smaller so it must be in $\langle a \rangle$.
Further in the proof he proves that the order of $b$ must be $p$ and this means $\langle b \rangle \cap \langle a \rangle = \{e\}$.
Right, same argument as above.
The induction hypothesis states that the decomposition into some $ \langle a' \rangle \times K'$ has already been proven for all groups of order $p^k$ less than $p^n$. Now the factor group $\overline{G} = G/\langle b \rangle$ does have order less than $p^n$ , so it can be decomposed this way by the assumption of the induction hypothesis for every $\overline a$ and some $\overline K$.
$\overline K$ consists of cosets of $\langle b \rangle$. The set $K$ is by definition all the elements of $G$ that are in these cosets $ \in \overline K $. So : $|K|=| \overline K| \cdot |\langle b \rangle|$ .
By Lagrange we have : $|G|=|\overline G| \cdot |\langle b \rangle| \implies |G|= |\langle \overline a \rangle | \cdot | \overline K | \cdot |\langle b \rangle|$
( last step is because : $ \overline G = \langle \overline a \rangle \overline K \land \langle \overline a \rangle \cap \langle \overline K \rangle = \{e\} \implies | \overline G|= |\langle \overline a \rangle | \cdot | \overline K | $ )
Because it is also proven above that : $|a|=|\overline a| \implies |\langle a \rangle |=| \langle \overline a \rangle |$ we have : $|G|= |\langle \overline a \rangle | \cdot | \overline K | \cdot |\langle b \rangle| = |\langle \overline a \rangle | \cdot | K | = |\langle a \rangle | \cdot | K | = |\langle a \rangle K |$
( last step because it is proven above that : $ \langle a \rangle \cap \langle K \rangle = \{e\}$ ) .
We have : $|G|=|\langle a \rangle K|$ $ \implies $ $\langle a \rangle K$ contains the same number of elements as $G $ $ \implies $ both sets must be equal.
With this last result we conclude that all the necessary conditions for $ \langle a \rangle \times K $ to be a valid internal direct product are fulfilled .