Discussion
Let G be a cyclic group of order 20. Then the number of elements of G of order 10 is :
Answer: B
There are 4 elements of order 10.
A cyclic group of order \(n\) has exactly \(\varphi(d)\) elements of order \(d\) for each divisor \(d\) of \(n\). Here \(d = 10\) divides 20, and
\[ \begin{aligned} \varphi(10) &= 10\left(1 - \tfrac{1}{2}\right)\left(1 - \tfrac{1}{5}\right) \\ &= 4 \end{aligned} \]
Writing \(G = \langle g \rangle\), they are \(g^{2}, g^{6}, g^{14}, g^{18}\), the powers \(g^{k}\) with \(\gcd(k, 20) = 2\).
10 is the order itself, not the number of elements having it. The elements of order 10 are exactly the generators of the unique subgroup of order 10, and a cyclic group of order 10 has \(\varphi(10) = 4\) generators, so 6 and 2 are wrong counts.
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