How can I find all elements of order $p^8$ in $G=\mathbb{Z}/p^3\mathbb{Z}\bigoplus\mathbb{Z}/p^5\mathbb{Z}\bigoplus\mathbb{Z}/p^7\mathbb{Z}\bigoplus\mathbb{Z}/p^9\mathbb{Z}\bigoplus\mathbb{Z}/p^{11}\mathbb{Z}$? I am wondering if I should use the Fundamental Theorem of Finitely Generated Abelian Groups, or the Chinese Remainder Theorem. Maybe I should use the fact that the order of any element in a direct sum is the least common multiple of the orders of its components? How can I figure out how many elements in $G$ have order = $\text{lcm}(\text{ord}(v),\text{ord}(w),\text{ord}(x),\text{ord}(y),\text{ord}(z))=p^8$?
Elements of order $p^8$ in direct sum
abstract-algebragroup-theory
Best Answer
Cyclic groups have a unique subgroup of every order dividing the order of the group.
The cyclic groups, $\Bbb Z_{p^9}$ and $\Bbb Z_{p^{11}}$, in this case, each have $\varphi(p^8)=p^8-p^7$ generators for the subgroup of order $p^8$.
We get $p^{15}p^8(p^8-p^7)+p^{15}(p^8-p^7)p^7=p^{15}(p^{16}-p^{14})=p^{29}(p^2-1)$, by taking products.
About the $p^7$ factor that appears, it's because there were still $p^8-\varphi(p^8)=p^7$ elements in the subgroup of order $p^8$ of $\Bbb Z_{p^9}$ that hadn't been used.
An easier way is to take the number of elements of order less than or equal to $p^8$ and subtract out those of order less than $p^8$. Sure enough, you get $p^{31}-p^{29}$.