Let $G$ be a finite abelian group and let $p$ be a prime that divides order of $G$. Then $G$ has at least an element of order $p$. Proof: I found this proof but I still struggling to understand it:
Answer: By induction over $|G|.$


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*if $|G|=2$, $1 \in G$ and G there is one more element $g\neq1$. Then $g=g^{-1}$ otherwise $g^{-1}$ should belong to $G$, because $G$ is group and $|G|=2$. Thus $g^2=1$, therefore $G$ there is an element $g$ wich has order $2$, and $2$ is prime.

*if $|g|>2$ and $p$ a prime such that $p$ divides $|G|$. Lets consider $|G|=mp$, $m$ a positive integer. If $1\neq g \in G$ then $o(g^m)=\frac{mp}{\gcd(m, mp)}=p$; Now, assuming that $p$ does not divide $o(g)=k$, then 
(now it starts my questions)


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*$p$ divides the index $|G:\langle g\rangle|$ (why I have this hypothesis, is this unique?)

*and, by induction, because $|G/\langle g\rangle|=|G|/k<|G|$, exists $x\langle g\rangle\in G/\langle g\rangle$ which has order $p$ (why?) in the quotient.
-So, setting $h=o(x)$ then $(x\langle g\rangle)^{h}=x^{h}\langle g\rangle=1\langle g\rangle=\langle g\rangle$ therefore $p$ divides $h$ and $o(x^{h/p})=p.$ (why?)
I look forward to seeing your answer.
 A: This is a special case of Cauchy's Theorem. A proof is given in this Wikipedia article.
A: A friend explained to me, he is very good at teaching, so I will share here
Question: Let $G$ be a finite abelian group and let $p$ be a prime that divides order of $G$. Then $G$ has at least an element of order $p$.
Proof (step-by-step):
By (strong) induction over $|G|$:

*

*If $|G|=2$, $G$ has 1 and one more element $g\neq 1$. So, $g^{-1}=g$, otherwise $g^{-1} \in G$ but $|G|=2$. Thus $g^{2}=1$, then $G$ has an element of order 2 and 2 is a prime number.


*If $|G|>2$ and $p$ is a prime number such that $p$ divides $|G|$, if $g \in G$ with $g\neq 1$ and $o(g)=mp$, for some $m \in \mathbb{Z}$, then $g^{mp}=(g^m)^p=1 \implies o(g^m)=p$.
Now, suppose that $p$ does not divide $o(g)=k$, and consider the group $\langle g \rangle$. By Lagrange theorem we have $|G|= \frac{|G|}{\langle g \rangle} \cdot \langle g \rangle$, so because $p$ does not divide $o(g)=k \implies p$ divides $|G:\langle g \rangle|$. By induction hypotesis because it holds for every subgroups that have order smaller than G, and because we have $|G:\langle g \rangle|=|G|/k< |G|$, it follows that there is $x\langle g \rangle \in G/\langle g \rangle$ wich has order $p$. Thus, considering $h=o(x)$ we have $(x\langle g \rangle)^h = x^{h}\langle g \rangle = \langle g \rangle.$ Then $p$ divides $h$, therefore $$((x^{\frac{h}{p}}))^{p} = x^h=1 \implies o(x^{\frac{h}{p}})=p.$$
