Consequence of Lemma: If G is abelian with exponent n, then $|G|\big\vert n^m$ for some $m\in N$

Question

Lemma: If G is abelian with exponent n, then $|G|\big\vert n^m$ for some $m\in N$. Theorem to be proved: Suppose G is finite abelian and group of order m, let p be a prime number dividing m. Then G has a subgroup of order p. My question is how to see that from the lemma that there is an element $x\in G$ which has divisible by p? I only see that there is an element of order $p\times m$ which is the exponent of the |G|=$p\times m$.

Answer 1

From the lemma (and the fact that $p$ is prime) we can deduce that $p$ must divide the exponent of $G$.

To obtain a contradiction, suppose that $g^p=1$ implies that $g=e$, i.e., that there is no element of order $p$. By what was said above, we may write $\exp (G)=pn$, so that for all $g\in G$, we have $1=g^{pn}=(g^n)^p$. By assumption, it follows that $g^n=1$ for all $g$, but this is a contradiction as $n<\exp(G)$. Hence, there must exist some element of order $p$ which generates a subgroup of order $p$.