Any finite non-cyclic abelian $p$-group will have more than $p-1$ elements of order $p.$
This question came into my mind while I was discussing to Shaun about a question I have just posted here. Any help regarding this will be highly appreciated.
Thank you very much.
Using the basic theorem you can write $G=C_1\times C_2\times\dots$ where $C_i$ are cyclic abelian $p$-groups. Since $G$ is not cyclic, there are at least two factors. Each factor has at least $p-1$ elements of order $p$. [If a generator is $a$ and the order is $p^n$, then take powers of $a^{p^{n-1}}$.]
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From the comment below, you don't want to use the structure theorem.
Take $C$ to be a cyclic subgroup of maximum possible order. As above, there are certainly at least $p-1$ elements of order $p$ in $C$.
Now take an element $x\in G\setminus C$ of smallest possible order. $x^p$ has smaller order than $x$, so $x^p\in C$. If $x^p$ generated $C$, then $x$ would be a larger cyclic subgroup than $C$ (contradicting the maximality of $C$). So $x^p$ must be a $p$th power in $C$, ie we can find $y\in C$ st $x^p=y^p$. Hence $xy^{-1}$ has order $p$ and does not belong to $C$. That gives us at least $p-1$ elements order $p$ outside $C$.