# Number of elements of given order in a group

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?

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$C_{20} \times C_{30} \cong C_4 \times C_5 \times C_2 \times C_3 \times C_5$

Yes, $C_{2}$ and $C_4$ each have subgroups of order 2:

$\varphi(4) = 2$, $\varphi(2) = 1$

"So number of elements of order 2 will be $(2 + 1)^2 - 1 = 3$, which was the correct answer."

$3$ is the correct number of subgroups of order $2$, but $3 = 2 + 1 \ne (2+1)^2 - 1 = 9 - 1 = 8$.

$\text{Aut}(C_{6125}) \cong \text{Aut}(C_{5^3}) \times \text{Aut}(C_{7^2})\not \cong C_{6125} \cong C_{5^3} \times C_{7^2}$

The automorphism group of a group is defined as a group whose elements are all the automorphisms of the base group (base group here $C_{6125}$)and where the group operation is composition of automorphisms. In other words, it gets a group structure as a subgroup of the group of all permutations of the group.

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 $\text{Aut}(C_{6125})$?

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How do I answer this question then? So how do you get that the number of subgroups is 3? Is it just 2 + 1? I had another question where I had to work out the number of elements of order 5 in the group $C_{42} \times C_{30}$ which after putting it in its primary decompositions, I got to be in the group $C_5 \times C_5$, and this gave me the number of elements as $(\varphi(5) + \varphi(1))^2 - \varphi(1)$ = 24, which was correct. – Kaish Dec 28 '12 at 22:33
How did you get $(2 + 1)^2 - 1 = 3$? – amWhy Dec 28 '12 at 22:49
Fortunate mistake I think. I think I saw there was a plus 1 and minus 1 so cancelled them both, then did $2^2 - 1$. – Kaish Dec 28 '12 at 22:52
It's always nice to encounter fortunate mistakes, as opposed to unfortunate ones, when one makes a mistake! – amWhy Dec 28 '12 at 22:56
and I believe you, @amWhy. Thanks for clearing that out and sorry that I asked. I've erased my comment. It's just that I was going to sleep and just passed by the puter and was greatly surprised to see the astonishing ressemblance of both answers' last lines. That's all. – DonAntonio Dec 29 '12 at 0:28

For the question on the number of elements of order two in $\,C_{20}\times C_{30}\,$ : each of both factors has one single element of order two (why?) , say $\,a\in C_{20}\,,\,b\in C_{30}\,$ . Thus, the elements of order two are $\,(a,1)\,,\,(1,b)\,,\,(a,b)\,$ . Can you see why there are no more?

As for the second question: since

$$6,125=7^2\cdot 5^3\Longrightarrow\left|Aut \left(C_{6,125}\right)\right|=\left|Aut(C_{49})\times Aut(C_{125})\right|=\left(7\cdot 6\right)(5^2\cdot 4)=4,200$$

There is one single element of order $\,7\,$ in $\,Aut(C_{49})\,$ and one single element of order $\,5\,$ and one of order $\,25\,$ in $\,Aut(C_{125})\,$ , say $\,a\,,\,b\,,\,c\,$ resp., so the elements of order $\,35\,$ are

$$(a^i,b^j)\,,\,(a^i,c^{5j})\;\;,\;\;1\leq i\leq 6\,\,,\,\,1\leq j\leq 4$$

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