Let G be a cyclic group of order 42. How to find the number of the elements of order 6, and the number of elements of order 7?

Question

Let $G$ be a cyclic group of order 42. How to find the nos of elements of order 6?

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score 2 · Answer 1 · answered Nov 23 '19 at 12:34

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Since $G$ is cyclic. It has a unique cyclic subgroup of order $d$, for each $d|42$.

If $a,b\in G$ bot have order $7$, they generate a cyclic subgroup of order $7$, that means that $(a)=(b)$ by uniqueness. Now, any element except the identity in this subgroup must have order $7$, since $7$ is prime (by an identical argument to the one below). So there are $6$ elements of order $7$.

On the other hand, consider the cyclic subgroup of order $6$. A cyclic group of order $6$ has $\varphi(6)=2$ generators (where $\varphi$ is Euler's Totient function). Since all cyclic groups are isomorphic, that means the cyclic subgroup also must have $2$ generators.

answered Nov 23 '19 at 12:34

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All this is reflected in the identity $n = \sum_{d \mid n} \phi(d)$. – lhf Nov 23 '19 at 12:38
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@lhf Your identity blown up my mind. – peterh Nov 23 '19 at 15:06
@peterh, see also https://math.stackexchange.com/a/190560/589 – lhf Nov 23 '19 at 18:27

Let G be a cyclic group of order 42. How to find the number of the elements of order 6, and the number of elements of order 7?

1 Answers1