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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?

2 Answers 2

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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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    Kaish - Have you been able to come up with an answer? $(a^i,b^j),\;(a^i,c^{5j})\;\;\;1\leq i\leq 6,\;\;1\;\leq j\;\leq 4$: for $(a^i,b^j)$, $i$ ranges from 1 to 6, and j ranges from 1 to 4, so there are $6\times4$ elements (ordered pairs) satisfying this possibility. For $(a^i, c^{5j})$, there are 6 possibilities for i, and $4$ possibilities for j, giving another 24 elements of order 35: 24 + 24 =$48$in all. That is there are $48$ elements of order 35.2013-01-09
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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

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