If d is a positive divisor of n, the number of elements of order d in a cyclic group of order n is $\phi \left ( d \right )$ where $\phi$ is the euler-phi function defined as the number of elements less than d and relatively prime to d. This by definition is $gcd\left ( d,n\right ) \equiv \phi\left ( d \right )$

The author provides the proof as follows. Proof: By the Fundamental theorem of cyclic group, there exists exactly one subgroup of order d-say, $\left \langle a \right \rangle$.

Then, every element of order d also generates the subgroup $\left \langle a \right \rangle$

The proof continues.

There is an underlying subtlety with the bold. I am unsure if my understanding of the bold is correct.

Note: $\left | a \right |=\left | \left \langle a \right \rangle \right |=d$.

This implies $a=\left \langle a \right \rangle$. Thus, a has order d and any element of order d generates $\left \langle a \right \rangle$

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    $\begingroup$ $a=\langle a\rangle$ does not make sense, since $a$ is an element and $\langle a\rangle$ is a set. And we already know $a\in\langle a\rangle$ since $a$ is a fixed element. We don't want to talk about just one element $a$, we want to talk about all elements $g$ of order $d$. So: suppose $g$ has order $d$, then we want to deduce $g$ is in $\langle a\rangle$. $\endgroup$ Feb 28, 2017 at 5:46

1 Answer 1


If $C$ is the unique subgroup of order $d$ and $g$ is any element of order $d$, then $|\langle g\rangle|=|C|$ implies $\langle g\rangle=C$ implies $g\in C$. If this is what you mean, then yes.

  • $\begingroup$ I shall continue with the given proof. An element $a^{k}$ generates the unique subgroup C of and only if gcd(k,d)=1. Thus, the number of such elements of order d in a cyclic group is $\phi(d)$ Can you shed some insight as to how this part of the proof is necessary? I do not see how it ties with the conclusion. $\endgroup$ Feb 28, 2017 at 6:33
  • $\begingroup$ @Mathematicing All elements of $C$ are of the form $a^k$ for some $1\le k\le d$. The generators (elements of order $d$) are precisely those with $\gcd(k,d)=1$. Thus, they are equal in number to the set of integers $1\le k\le d$ with $\gcd(k,d)=1$. That number is, by definition, $\phi(d)$. $\endgroup$ Feb 28, 2017 at 6:47
  • $\begingroup$ I got it. I got the notation mixed up. $\endgroup$ Feb 28, 2017 at 6:50

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