How many elements of order $d$ are there in $\mathbf{Z}_{10}\times \mathbf{Z}_{10}$ where $d$ is a divisor of $10$? I completed exercise $6$ page $166$ in Dummit and Foote, Abstract Algebra.  It reads: "Let $G$ be a finite abelian group of type $(n_{1}, n_{2}, ..., n_{s})$. Prove $G$ contains an element of order $m$ if and only if $m$ divides $n_{1}$."
This exercise made me wonder if there is anything at all that we can say about the number of elements of each permissable order in a non-cyclic abelian group like $\mathbf{Z}_{10}\times \mathbf{Z}_{10}$. 
I wrote a code in Mathematica to find that there is 1 element of order 1, 3 elements of order 2, 24 elements of order 5, and 72 elements of order 10.
 A: Using Chinese Remainder theorem you get :
$$\mathbb{Z}/10\mathbb{Z}\times \mathbb{Z}/10\mathbb{Z}\text{ is isomorphic to }[\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}]\times[\mathbb{Z}/5\mathbb{Z}\times \mathbb{Z}/5\mathbb{Z}]  $$
Now write :
$$S_2:=\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z} $$
$$S_2:=\mathbb{Z}/5\mathbb{Z}\times \mathbb{Z}/5\mathbb{Z}$$
Now your group $G$ is written as $S_2\times S_5$ if $g=(s,t)\in S_2\times S_5$ then $o(g)=o(s)o(t)$ since $s$ and $t$ have order coprime to each other.
Now in $S_2$ you have one element of order $1$ and three elements of order $2$. In $S_5$ you have one element of order $1$ and twenty-four elements of order $5$. 
Finally you get :
$$\text{ one element of order } 1 \text{ of the form } (0,0) $$
$$\text{ three elements of order } 2 \text{ of the form } (s,0)\text{ where } s\neq 0 $$
$$\text{ twenty-four elements of order } 5 \text{ of the form } (0,t)\text{ where } t\neq 0 $$
$$3\times 24=\text{ seventy-two elements of order } 10\text{ of the form } (s,t)\text{ where } s,t\neq 0 $$
