Let $G$ be an abelian group of order $p^m$. Suppose that $G$ is a direct sum of $t$ cyclic groups. Show that the subgroup $H$ of $G$ containing $0$ and order $p$ elements has order $p^t$.
By the structure theorem of finite abelian groups, we have \begin{equation} G=\mathbb{Z}a_1 \oplus … \oplus \mathbb{Z}a_t \end{equation} where each $a_i$ has order $p^{\alpha_i}$ and $\alpha_1+…+\alpha_t=m$. Since direct summands are cyclic, each of their elements is an integral multiple of some fixed $a_i$ in $G$. So every $x\in G$ can be written as \begin{equation} x=\sum_{i=1}^tn_ia_i, \end{equation} and the order of $x$ is the least common multiple of the orders of elements $n_ia_i$. At this moment, I don’t know how to find the order of $H$. What does an element of order $p$ looks like in terms of the $\mathbb{Z}a_i$ ? Thanks for any hint/help.
