I have this Lemma: If $\displaystyle G$ is a finite abelian group whose order is divisible by a prime $\displaystyle p$, then $\displaystyle G $ contains an element of order $\displaystyle p$.

I have shown that for $\displaystyle x \in G$, $\displaystyle x \ne \{{1}\}$ and order of $\displaystyle x$ is $\displaystyle pm$, then $\displaystyle x^{m}$ is an element in $\displaystyle G$ of order $\displaystyle p$.

Now, suppose that $\displaystyle x$ has order $\displaystyle t$, where $\displaystyle (p,t)=1$. Since $\displaystyle G$ is abelian, $\displaystyle <x>$ is a normal subgroup of G and $\displaystyle G/<x>$ is an abelian group of order $\displaystyle |G|/t$. It is easy to see that $\displaystyle |G|/t$ is divisible by $\displaystyle p$ (since $\displaystyle |G|$ is divisible by $\displaystyle p$ and $\displaystyle (p,t)=1$) and is less than $\displaystyle |G|$.

Afterwards I know I need to show that there is an element in $\displaystyle G/<x>$ whose order is $\displaystyle p$, but I'm stuck here. Please help me in completing the proof.

I have another corollary: A finite group $\displaystyle G$ is a $\displaystyle p$-group $\displaystyle \Longleftrightarrow$ $\displaystyle |G|$ is a power of $\displaystyle p$.

I have shown $\displaystyle ( \Longrightarrow )$.

For $\displaystyle ( \Longleftarrow )$, I used Lagrange's Theorem and shown that every subgroup of $\displaystyle G$ has order of power of $\displaystyle p$. How to continue the proof from here?