# Number of elements of each order in a cyclic group

Theorem:

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 $$\left \langle a \right \rangle$$

• $a=\langle a\rangle$ 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 $\langle a\rangle$. Feb 28, 2017 at 5:46

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.
• 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. Feb 28, 2017 at 6:33
• @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, $\phi(d)$. Feb 28, 2017 at 6:47