When $d \mid n$, the number of elements of order $d$ in a cyclic subgroup of order $n$ is $\phi (d)$

Question

If $d$ is a positive divisor of $n$, the number of elements of order $d$ in a cyclic subgroup of order $n$ is $\varphi (d)$

where $\varphi(d)$ is the Euler's totient function, denoting the number of positive numbers less than $d$ that are coprime to $d$.

The question I have concerns a part of the proof:

If $d \mid n$ then there exists exactly one subgroup of order $d$ , call it $\langle a \rangle$. Then every element of order $d$ also generates the subgroup $\langle a \rangle$ and an element $a^k$ generates $\langle a \rangle$ iff $\gcd(k,d) = 1$ implies that the number of such elements is precisely $\varphi (d)$.

How does every element of order $d$ also generate the subgroup $\langle a \rangle$, wouldn't it be only one $a$ since $|\langle a \rangle| = |a|$? And how does this fact imply $\varphi(d)$ is the correct number?

Answer 1

Let $G=\langle a\rangle $ is a finite cyclic group of order $n$ and $d\mid n$.

Then $G$ has a unique subgroup of order $d$ say $H_d$ and $H_d=\langle a^{\frac{n}{d}}\rangle$ ( see here for proof).

Now number of elements of order $d$ are precisely the number of generators of $H_d$.

$a^{r}\in H_d$ generates $H_d$ iff $1\le r\le d$ and $\gcd(r, d) =1$

$($ $|a^r|=\frac{d}{\gcd(r, d) }$ . Then $|a^r|=d$ iff $\gcd(r, d) =1$$)$

Hence the number of generators of $H_d$ is exactly $\varphi(d) $ which are precisely the elements of order $d$.

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Given $G=\langle a\rangle$, $|G|=n$ and $d\mid n$, show $G$ has a unique subgroup of order $d$.

Answer 2

Let $G=\langle a\rangle $ is a finite cyclic group of order $n$ and $n=dk$.

$a^n=e\Rightarrow a^{kd}=e$, define $a^k=b\Rightarrow b^d=e$, so we can find a unique subgroup with order $d$.

Define: $H=\langle b\rangle$, now this problem is converted to: how many generators to generate $H$?

One natural choice is $\langle b\rangle$, then how to find other generators of $H$, such that $\langle b^s\rangle=\langle b\rangle$?

So how many choices for $s$? ($1\leq s\leq |\langle b \rangle|$)

Note that $|\langle b\rangle|=d$, and we have $\langle b^s\rangle=\langle b^{\gcd(s, d)}\rangle$, hence

$d=|\langle b\rangle|=|\langle b^s\rangle|=|\langle b^{\gcd(s, d)}\rangle|=\frac{d}{\gcd(s, d)}\Rightarrow \gcd(s, d)=1$

So we need to count the number of $s$ such as $\gcd(s, d)=1$, which is exactly equal to $\phi(d)$.

Answer 3

An element of order d generated a subgroup of order d which is $<a>$ since $<a>$ is unique.

A subgroup of order $d$ is isomorphic to $Z/d$ if $[n]$ generates $Z/d$, $[1]=m[n]$ this is equivalent to saying $1=mn+cd$, thus $gcd(d,n)=1$. Thus there exists $\phi(d)$ generators.

Answer 4

Consider $\mathbb{Z}_5$. Under addition we have $$\mathbb{Z}_5=\langle 1 \rangle=\langle 2 \rangle=\langle 3 \rangle=\langle 4 \rangle$$

A cyclic group has at least one generator, but if it is finite, then it will have exactly $\phi (n)$ total, where $G$ is your cyclic group and $n=\left| G \right|$. So equivalently, you could define the finite cyclic group of order $n$ as the group with exactly $\phi(n)$ generators, but it would need to be proven that this is equivalent to being generated by a single element.