Example 1: "Calculate the number of elements of order 2 in the group $C_{20} \times C_{30}$"
To do this, I split the groups into their primary decompositions and got that the groups with elements of order 2 are $C_4$ and $C_2$. From here, to then work out the number of elements of order 2 I did:
$\varphi(4) = 2$,
$\varphi(2) = 1$
So number of elements of order 2 will be $(2 + 1)^2 – 1 = 3$, which was the correct answer.
However
Example 2: "Compute the number of elements of order 35 of the group $\mathrm{Aut}(C_{6125})$"
To do this, I can just check that 35 divides 6125 and then use the Euler totient function. Why do I not have to split 6125 into its primary decomposition and then use that little formula to work out the number of elements? Is it because this is a cyclic group and so I can just use the Euler function, however as the other one is a direct product, I need to use a different method?
Best Answer
$C_{20} \times C_{30} \cong C_4 \times C_5 \times C_2 \times C_3 \times C_5$
Yes, $C_{2}$ and $C_4$ each have subgroups of order 2:
$\varphi(4) = 2$, $\varphi(2) = 1$
$3$ is the correct number of subgroups of order $2$, but $3 = 2 + 1 \ne (2+1)^2 - 1 = 9 - 1 = 8$.
$\text{Aut}(C_{6125}) \cong \text{Aut}(C_{5^3}) \times \text{Aut}(C_{7^2})\not \cong C_{6125} \cong C_{5^3} \times C_{7^2}$
There is exactly one element of order $\,7\,$ in $\,\text{Aut}(C_{7^2})\,$ and exactly one element of order $\,5\,$ and exactly one of order $\,25\,$ in $\,\text{Aut}(C_{5^3})\,$, let's call them $\,a,\,b,\,c\,$ respectively. Then the elements of order $\,35\,$ are as follows:
$$(a^i,b^j),\;(a^i,c^{5j})\;\;\;1\leq i\leq 6,\;\;1\;\leq j\;\leq 4$$
Can you compute the number of elements of order $35$ in $\text{Aut}(C_{6125})$?