Given a group, say:
Z2 × Z3 × Z3 × Z9 × Z27 × Z5.
How can I find the number of elements of each prime power order?
For order 1, the number of elements is 1 (the identity).
For order 2 and above, I am completely lost and my textbook is not much help at all. A simple explanation would be so appreciated!
Elements of order $2$ can only be in $Z_2$ and so there is $1$.
Elements of order $5$ can only be in $Z_5$ and so there are $4$.
The powers of $3$ require slightly more work. For example, elements of order $3$ are in $$Z_3 \times Z_3 \times Z_9 \times Z_{27}$$ and therefore in $$Z_3 \times Z_3 \times Z_3 \times Z_{3}$$ and there are therefore $3^4-1$ of these.
Can you do elements of $9$ and $27$ yourself? Ask if you want your answers checked and remember that every element of $$Z_3 \times Z_3 \times Z_9 \times Z_{27}$$ will have order a power of $3$ and that there are $3 \times 3 \times9 \times {27}$ of these elements in total.