## Gallian- Ch. 8 #11

Problem Statement: How many elements of order $4$ does $\mathbb{Z}_4\oplus\mathbb{Z}_4$ have? Explain why $\mathbb{Z}_4\oplus\mathbb{Z}_4$ has the same number of elements of order $4$ as $\mathbb{Z}_{8000000}\oplus\mathbb{Z}_{400000}$. Generalize to the case of $\mathbb{Z}_m\oplus\mathbb{Z}_n$.

Solution: First let’s find how many elements there are of order $4$ in $\mathbb{Z}_4\oplus\mathbb{Z}_4$. An element $(a, b)\in\mathbb{Z}_4\oplus\mathbb{Z}_4$ has order $4$ if $lcm(\left|a\right|,\left|b\right|)=4$. This results in five possible cases we need to consider:

First: If $\left|a\right|=1, \left|b\right|=4$. Then there is one choice for $a$ since $\phi(1)=1$ and there are two choices for $b$ since $\phi(4)=2$. So we have a total of $2$ possible $(a,b)$.

Second: If $\left|a\right|=4, \left|b\right|=4$. Then there are two choices for $a$ and there are two choices for $b$ since $\phi(4)=2$. So we have a total of $4$ possible $(a,b)$.

Third: If $\left|a\right|=2, \left|b\right|=4$. Then there is one choice for $a$ since $\phi(1)=1$ and there are two choices for $b$ since $\phi(4)=2$. So we have a total of $2$ possible $(a,b)$.

Fourth: If $\left|a\right|=4, \left|b\right|=2$. Then there are two choices for $a$ and there is one choice for $b$. So we have a total of $2$ possible $(a,b)$.

Fifth: If $\left|a\right|=4, \left|b\right|=1$. Then there are two choices for $a$ and one choice for $b$. So we have a total of $2$ possible $(a,b)$.

Now, adding up all these possible $(a,b)$ we see that there are $12$ elements of order $4$ in $\mathbb{Z}_4\oplus\mathbb{Z}_4$.

$\mathbb{Z}_4\oplus\mathbb{Z}_4$ has the same number of elements of order $4$ as $\mathbb{Z}_{8000000}\oplus\mathbb{Z}_{400000}$ since the only thing that determines how many elements there are of each order are the divisors of $4$. Also note that it is necessary that the divisors also divide the orders of the groups in the direct product. Since $4$ divides $4, 8000000,$ and $400000$ it follows that all of the divisors of $4$ also divide those numbers and so the process outlined above will be the exact same for $\mathbb{Z}_{8000000}\oplus\mathbb{Z}_{400000}$.

In general, if you are trying to count the number of elements of order $d$ in $\mathbb{Z}_m\oplus\mathbb{Z}_n$ the result will be the same for any $m$ and $n$ so long as $d$ divides both $m$ and $n$.

Reflection: One more key thing to note here is that this only works when we are dealing with cyclic groups. Another way one could approach this problem is by writing out the elements of the group. Since we are only dealing with $16$ elements it’s not that hard to do, but for larger groups, like $\mathbb{Z}_{8000000}\oplus\mathbb{Z}_{400000}$, it is much more difficult. This method is also helpful when trying to count cyclic subgroups of order $d$. Suppose we wanted to know how many cyclic subgroups there are in $\mathbb{Z}_4\oplus\mathbb{Z}_4$ with order $4$. First we would find how many elements there are of order $4$, we’ve shown above that there are $12$, and then we would divide that number by $\phi(4)=2$. In general if you want to know how many cyclic subgroups there are of order $d$ you first find how many elements there are of order $d$ and then divide that number by $\phi(d)$. Again, you could do this by writing out all of the elements, but that is more time consuming for large groups.