Order of Cyclic Subgroups

Let $G$ be a cyclic group with $n$ elements be generated by $a$. Then call the cyclic subgroup $\langle a^{d} \rangle = H$. My professor says that $|H| = \frac{n}{d}$ while the book says the $|H| = \frac{n}{gcd(n, d)}$. Are they the same or am I very confused?

Thanks in advance.

The book is correct; your professor is correct when $d$ divides (i.e. goes into) $n$.

Note that for any $g\in G$, $$|\langle g\rangle|=\text{ord}(g).$$ The general relation is that, if $\text{ord}(a)=n$, then $$\text{ord}(a^d)=\frac{n}{\gcd(d,n)}.$$ For example, consider the cyclic group $G=\mathbb{Z}/6\mathbb{Z}=\{0,1,2,3,4,5\}$ with operation $+\,$, and let $a=1$. It has order $n=6$. Let $d=5$; then $a^5$ means $$a+a+a+a+a=5$$ so $H=\langle a^5\rangle=\langle 5\rangle=G$, so $|H|=6=\frac{6}{\gcd(5,6)}$. In contrast, the statement that $|H|=\frac{6}{5}$ doesn't even make any sense.

If you intended $n$ to be the order of $G$, then both answers will be false unless $G$ is a cyclic group and $\langle a\rangle=G$ (in my answer above, I assumed you meant $n=\text{ord}(a)$, which agrees in this case, i.e. $\text{ord}(a)=n=|G|$ if and only if $G$ is cyclic and $G=\langle a\rangle$). For example, in the group $$G=(\mathbb{Z}/6\mathbb{Z})\times(\mathbb{Z}/2\mathbb{Z}),$$ for which $|G|=12$, the element $a=(1,1)$ has order 6, and $$a^5=(1,1)+(1,1)+(1,1)+(1,1)+(1,1)=(5,1),$$ hence $$H=\langle a^5\rangle=\langle(5,1)\rangle$$ has $|H|=6$, but $$|H|\neq\frac{12}{1}=\frac{12}{\gcd(5,12)}$$ and $$|H|\neq\frac{12}{5}$$

The book is correct - it is the statement of the Fundamental Theorem of Cyclic Groups. Its proof is rather simple:

Let $t$ belong to <$a^d$>, then $t$ = $a^{dq}$, where $q$ is an integer. Let $s = \gcd(n,d)$. Then $d = sp$, for some integer $p$. Then, $t = a^{spq},$ so $t= (a^s)^{pq}$, so $t$ belongs to <$a^s$> = <$a^\gcd(n,d)$>. Thus, <$a^d$> is a subset of <$a^{\gcd(n,d)}$>.

Let $r$ belong to <$a^{\gcd(n,d)}$>, then $r = a^{sm}$, for some integer $m$. Now, by Bezout's identity, there exist $x,y \in \mathbb{Z}$ such that $\gcd(n,d)= s = nx + dy$, so $r= a^{nx + dy}= a^{dy}$, since $a^{nx} =e$. But, $a^{dy}$ belongs to <$a^d$>, so, we are done, that is, we have proved that <$a^{\gcd(n,d)}$> is a subset of <$a^d$>, so from our results, we have: <$a^s$> = <$a^d$>, where $s = \gcd(n,d)$.