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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)?

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What if the order of $G$ is $np^k$? – John Habert Mar 25 '14 at 16:35
So that means we could have direct sums $C_n \oplus C_{p} \oplus C_{p}$, and $C_n \oplus C_{p} \oplus C_{p} \oplus C_{p}$, and all the way up to $C_n \oplus C_{p} \oplus C_{p} \oplus \cdots \oplus C_{p}$ ($k$ terms of $C_{p}$)? – Cookie Mar 25 '14 at 16:38
Not necessarily. $C_p \oplus C_p \not\simeq C_{p^2}$ as the latter has elements of order $p^2$ while the former does not. – John Habert Mar 25 '14 at 16:43
So the only direct sum that I find (that is isomorphic to $G=C_{np^k}$) is $C_n \oplus C_{p^k}$. – Cookie Mar 25 '14 at 17:04
That is not true though. Think about $(\mathbb{Z}/2 \mathbb{Z})^2$ and $\mathbb{Z}/ 4 \mathbb{Z}$ one has an element of order 4, one does not. – Alexander Mar 25 '14 at 17:46
up vote 2 down vote accepted

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

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Because $p^k = p^{k-1} p$, so that's why we use the new element $g^{p^{k-1}}$, right? – Cookie Mar 25 '14 at 20:14
yep! That is correct – Alexander Mar 26 '14 at 2:10

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

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