I am trying to verify the following proposition:
Let $G$ be a finite Abelian group and let $p$ be a prime that divides the order of $G$. Then $G$ has an element of order $p$.
My proof: By Lagrange's theorem: $x^{|G|}=e$. By assumption we have $kp=|G|$ for some prime $p$. So $e=x^{|G|}=x^{kp}$. Thus $|x^k|=p$. $\blacksquare$
The book's proof uses induction and cosets -- is this necessary?
For reference, here's the book proof:
Clearly, this statement is true for the case in which $G$ has order 2. We prove the theorem by using the Second Principle of Mathematical Induction on $|G|$. That is, we assume that the statement is true for all Abelian groups with fewer elements than G and use this assumption to show that the statement is true for G as well. Certainly, G has elements of prime order, for if $|x| = m$ and $m = qn$, where $q$ is prime, then $|x^n| = q$. So let $x$ be an element of $G$ of some prime order $q$, say. If $q=p$, we are finished; so assume that $q \neq p$. Since every subgroup of an Abelian group is normal, we may construct the factor group $\bar{G} = G/\langle x\rangle$. Then $\bar{G}$ is Abelian and $p$ divides $|G|$, since $|\bar{G}| = |G|/q$. By induction, then, $G$ has an element — call it $y\langle x\rangle$ — of order $p$. Then, $(y\langle x\rangle)^p = y^p\langle x\rangle = \langle x\rangle$ and therefore $y^p \in \langle x\rangle$. If $y^p = e$, we are done. If not, then $y^p$ has order $q$ and $y^q$ has order $p$. $\blacksquare$