I’m wondering if there is a counterexample of the following statement.
Let $G$ be a finite abelian group and $H$ be a subgroup of $G$. Suppose that $G$ has an element of order $p$ (prime). Let the positive integer $n$ be the number of elements of order $p$. Then $G/H$ has at most $n$ elements of order $p$.
I don’t know if it is true or not. I tried to prove it but I couldn’t
Can anyone help me to prove or disprove it?
Thank you
Writing $G$ as a direct sum of cyclic groups of prime power order, we see that $n$ is just $p^k-1$ where $k$ is the number of summands whose order is a power of $p$. This number $k$ can also be described as the dimension of $G/pG$ as a vector space over $\mathbb{F}_p$. Now let $K=G/H$ and observe that the quotient map $G\to K$ induces a surjective homomorphism $G/pG\to K/pK$, and thus $\dim K/pK\leq\dim G/pG$. That is, $K$ has at most $k$ summands whose order is a power of $p$, and thus also has at most $p^k-1$ elements of order $p$.