Abelian Groups Have An Element Of Order $p$ For Every Prime That Divides The Order Of The Group

Hint: This problem is done by induction on the order of $G$.

For each integer $k\geq 2$, let $P(k)$ be the statement, "For every positive integer $n<k$, if $|G|=n$ and $p$ divides $|G|$ then $G$ has an element of order $p$." Use induction to prove that $P(k)$ is a true statement for every integer $k\geq 2$. This will require you to do two things.

1. Is the statement $P(2)$ true? Is the statement $P(3)$ true?
2. If you assume for some specific $k\geq 2$ that the statement $P(k)$ is true, then prove that the statement $P(k+1)$ is true. To prove $P(k+1)$ is a true statement, you have to prove that if $n<k+1$ and $|G|$ has order $n$ with $p$ dividing $G$, then $G$ has an element of order $p$. This is quite a lot of facts to prove. However, the induction assumption should prove every single one of them instantly except for when $n=k$. So you must show that if $|G|=k$ and $p$ divides $|G|$, then $G$ has an element of order $p$. You know for sure that $G$ has an element of some order other than 1, say $m=|a|$, which means that it has an element $b$ of some prime order $q$ (why?). If $q=p$ we're done. Otherwise, consider the factor group $G/\left<b\right>$. What is the order of $G/\left<b\right>$? Does $p$ divide $G/\left<b\right>$? What does the induction assumption then tell you? How does problem 103 help?