What is the number of elements of order 3 in the internal direct product $C_3 \times C_9$ of $C_3$ and $C_9$ where $C_i$ is the cyclic group of order $i$.
My work so far, let $(a,b) \in C_3 \times C_9$ Then $(a,b)^3=(a^3,b^3)$.
Case: $a$ is of order 3(2 such elements). Then $b$ is of order 1 or 3, of which there are 3 such elements. So there are $2*3=6$ options.
Case: $a$ has order 1, so is the identity element. Then $b$ must be of order $3$ of which there are two elements. So $2$ options in total. So this would give 8 elements. Is this correct and also is there a more succinct way of writing this?
Yes, your answer is correct. Here are two slicker approaches to the problem:
Count the number of elements whose order divides $3$. This corresponds to elements of the form $(a,b)$ where $a$ and $b$ have order $1$ or $3$. There are three choices for each of $a$ and $b$, so this leads to $9$ elements. Of these elements, exactly one is the identity, with order $1$, so there are $8$ of order $3$.
The set of elements of $C_3\times C_9$ of order dividing $3$ forms a group (since the order of an element of an abelian group is the least common multiple of the orders of its factors). Since the order of $C_3\times C_9$ is $27$, by Lagrange's theorem, the order of this subgroup is $1$, $3$, $9$, or $27$. There are at least $4$ elements of order $3$ and there can't be $27$ since there are elements of order $9$. Hence, the order of the group must be $9$. Subtracting the identity gives $8$. Note that the group of interest is isomorphic to $C_3\times C_3$.