Proof of Fundamental Theorem of Finite Abelian Groups 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.  
 A: This is Lemma 2 Ch. 11 from : 'Gallian , Contemporary abstract algebra' leading to the fundamental theorem.

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 $p^{k-1}$?]

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 $n$.

Then $x^{p^m} = e$ for all $x$ in $G$. [Question: How, if $|G| = p^k$ ?]

$|G|=p^n$ , 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 : $x^{p^m}= (x^{p^j})^{p^{m-j}} = e^{p^{m-j}} =e$

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 $b^p = e$ ?] 

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 $\langle b \rangle \cap \langle a \rangle = \{e\}$.

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?]

Right, same argument as above.

By induction, we know that $\overline{G} = \langle \overline a \rangle \times \overline{K}$. [Question: ??? What induction? What is the basis of this step? What even is $K$? How are we even defining $K$ in the steps below?] 

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 $\overline K$.

It now follows from an order argument (???) that $G = \langle a \rangle K$, and therefore $G = \langle a \rangle \times K$.

$\overline K$ 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 .
A: With regards to Question 1: No, the converse is not true.  Consider the group $Z_2 \times Z_2$ - it has order 4 but no element of order 4.  The 'maximum m' issue revolves around cases like this.
With regards to the 'what induction?' issue: The entire argument is an inductive one - it is a 'course-of-values' induction where one shows that, if the hypothesis is true for all sizes smaller than N, then it is true for size N.  The gist of the argument here is as follows: Choose a cyclic subgroup A of maximum size m.  If this is the whole group, we're done [the base case].  Otherwise, we can find another cyclic subgroup B of order p whose intersection with A is the trivial subgroup.  We pick order p for B because we don't need a larger order for this argument, and we know we can find such a B for order p [not necessarily for larger orders].  Now we factor the original group by B to get a smaller group [this is our induction step]  We chose B to have trivial intersection with A so that the image of A in the factor group is isomorphic with A - i.e A doesn't shrink.  Since [by induction] A (or rather, a group isomorphic with A) is a factor of the smaller group, projecting back we find that A is a factor of the original group.
