# Elements of order $9$ in $C_{135} \times C_6 \times C_9$

Ok, so I first split everything up into powers of primes and get

$$C_{27} \times C_5 \times C_2 \times C_3 \times C_9$$

Now, I want to find elements of order 9 in each of these cyclic groups. This means I get:

• $\varphi(27) = 18$ in $C_{27}$
• None in $C_5$
• None in $C_2$
• Pick arbitrary $3$ for $C_3$
• $\varphi(9) = 6$ in $C_9$

And so the number of elements of order $9$ in my group is $18 \cdot 3 \cdot 6 = 324$. IS this correct?

-
The number of elements of order $9$ in a cyclic group does not depend on the order of that group other than to check whether $9$ divides the order (it will always be $6 = \varphi(9)$ as long as that is the case). Come to chat, it is slow anyway. – Tobias Kildetoft Jan 14 '13 at 14:53
Basically, you want elements of $C_{27}$ and $C_9$ whose maximum order is $9$. Then you can pick any element of $C_3$. The best way to do this is a simple inclusion-exclusion – Thomas Andrews Jan 14 '13 at 14:57
Also, there can be elements of order 9 in the product such that one component has order 9 and the other components have orders that are divisors of 9. – Andreas Blass Jan 14 '13 at 14:57
Oops. The "Also" in my previous comment referred to Tobias's comment. Thomas's comment, which he added while I was typing mine, already incorporates what I said. – Andreas Blass Jan 14 '13 at 14:58
@AndreasBlass Yeah, but yours is more general. :) – Thomas Andrews Jan 14 '13 at 14:59

For the total element $x = (x_{27}, x_5, x_2, x_3, x_9)$ to have order $9$, all of the $x_n$ need to have order a divisor of $9$, and at least one of them have to have order exactly $9$. In the case of $x_5$ and $x_2$, the only element it could be is the respective identity elements. So we need to pick:
1. An element $x_{27}\in C_{27}$ of order $1$ (one element), $3$ (two elements), or $9$ (six elements).
2. An element $x_3 \in C_3$ (three elements)
3. An element $x_9 \in C_9$ of order $1$ (one element), $3$ (two elements) or $9$ (six elements)
4. At least one of the two elements from 1. and 3. has to be of order $9$.
And there we have it. So we count: If $x_{27}$ has order $9$, then the rest can be chosen freely, and we end up with $6\times 3\times 9 = 162$ different elements. If $x_{27}$ does not have order $9$, then $x_9$ must be of order $9$, and we get $3 \times 3 \times 6 = 54$ different elements. In total we get $162 + 54 = 216$ elements of order $9$ in your group.
There are only three elements of order $1,3$ in $C_{27}$, I think, not $4$ – Thomas Andrews Jan 14 '13 at 15:20