[proofplan] Because $G$ is cyclic, it is abelian, so every subgroup of $G$ is normal and the quotient $G/H$ is a group. To prove that the quotient is cyclic, we express an arbitrary element of $G$ as a power of $g$ and pass to cosets. Finally, we count the quotient directly: the cosets of $H$ partition $G$, and each coset has exactly $d$ elements. [/proofplan] [step:Use cyclicity to make the quotient group available] Since $G=\langle g \rangle$, the group $G$ is cyclic. By [citetheorem:8243], $G$ is abelian. Therefore, for every $a \in G$ and every $h \in H$, we have $aha^{-1}=aa^{-1}h=h$, so $aHa^{-1}=H$. Hence $H \trianglelefteq G$, and the [quotient group](/theorems/790) $G/H$ is defined. [/step] [step:Show every coset is a power of $gH$] Let $xH \in G/H$ be an arbitrary coset. Since $G=\langle g \rangle$, there exists $k \in \mathbb{Z}$ such that $x=g^k$. In the quotient group $G/H$, the power of the coset $gH$ satisfies \begin{align*} (gH)^k=g^kH. \end{align*} Thus \begin{align*} xH=g^kH=(gH)^k. \end{align*} Since every element of $G/H$ is a power of $gH$, we obtain \begin{align*} G/H=\langle gH \rangle. \end{align*} [guided] We want to prove that $G/H$ is cyclic and identify a generator. The natural candidate is the coset $gH$, because $g$ generates $G$ itself. Let $xH \in G/H$ be arbitrary. Since $G=\langle g \rangle$, the definition of generated subgroup gives an integer $k \in \mathbb{Z}$ such that $x=g^k$. Passing from elements to cosets gives \begin{align*} xH=g^kH. \end{align*} In the quotient group, multiplication is defined by $(aH)(bH)=(ab)H$. Therefore integer powers are compatible with taking cosets, and for this integer $k$ we have \begin{align*} (gH)^k=g^kH. \end{align*} Consequently, \begin{align*} xH=(gH)^k. \end{align*} Because the coset $xH$ was arbitrary, every element of $G/H$ is a power of $gH$. Hence $gH$ generates the quotient: \begin{align*} G/H=\langle gH \rangle. \end{align*} [/guided] [/step] [step:Count cosets to compute the order of the quotient] Let $q:=|G/H|$ denote the number of cosets of $H$ in $G$. For a fixed element $a \in G$, define the map \begin{align*} L_a: H &\to aH \end{align*} by \begin{align*} L_a(h)=ah. \end{align*} The map $L_a$ is injective because $ah_1=ah_2$ implies $h_1=h_2$ after multiplying on the left by $a^{-1}$. It is surjective by the definition of the coset $aH$. Hence every coset $aH$ has exactly $|H|=d$ elements. The distinct cosets of $H$ partition $G$. Since there are $q$ cosets and each has $d$ elements, we get \begin{align*} |G|=qd. \end{align*} Using $|G|=n$, this becomes \begin{align*} n=qd. \end{align*} Therefore \begin{align*} |G/H|=q=\frac{n}{d}. \end{align*} This proves both asserted conclusions. [/step]