Using the Fundamental Theorem of Finite Abelian Groups

Question

Let $G$ be a finite abelian group and let $p$ be a prime that divides the order of $G$. Use the Fundamental Theorem of Finite Abelian Groups to show that $G$ contains an element of order $p$.

The Fundamental Theorem of Finite Abelian Groups states that a finite abelian group $G$ is the direct summation of cyclic groups, each one of them are of prime power order.

For this problem, if $p$ divides $G$, then $G$ must have an order of $np^k$, where $n$ and $k$ are positive integers. So applying the fundamental theorem, I get the direct sum $$C_n \oplus C_{p^k}$$ with elementary divisors of $n$ and $p$, and with an invariant factor of $np$ (so the direct sum is isomorphic to the group $C_{np}$. Seeing that there is $C_p$ from the direct sum above, does this suffice to show that $G$ contains an element of order $p$ (or a subgroup of order $p$ that is generated by the element)?

Answer 1

HINT: If $g \in G$ has order $p^n$, then $g^{p^{n-1}}$ has order $p$.

Answer 2

This isn't an answer, but I can't fit this properly in the comment section.

At this point, I have a group of prime power order, in particular $C_n \oplus C_{p^k}$. My textbook (Hungerford, 2nd edition) is suggesting as a hint to use Theorem 7.8, which says

Let $G$ be an additive group and let $a \in G$.

• If $a$ has infinite order, then the elements $a^k$, with $k \in \mathbb{Z}$, are all distinct.
• If $a^i = a^j$ with $i \not= j$, then $a$ has finite order.
• If $a$ has finite order $n$, then $$a^k=e \text{ if and only if } n \mid k$$ and $$a^i = a^j \text{ if and only if } i \equiv j \text{ (mod $n$)}$$
• If $a$ has order $n$ and $n = td$ with $d > 0$, then $a^t$ has order $d$.

...in order to find an element of order $p$. The fourth bullet helps finish the job.