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

Question

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

Proof:

(Existence) : $|\langle a\rangle|=|a|=n$ and $ |a^\frac{n}{d}|=d$. Then, $H_d=\langle a^\frac{n}{d}\rangle$ is a subgroup of $G$ of order $d$.

(Uniqueness) :

Suppose, $H\le G$ and $|H|=d$.

Claim: $H=H_d$.

It is enough to prove one-sided set inclusion $H\subseteq H_d$ because $H\le H_d$ and $|H|=|H_d|$ implies $H=H_d$.

Choose $b\in H$ . Then $|H|=d \implies b^d=e$ and $b\in H \implies b \in G$.

Hence $b=a^k$ for some $k\in \mathbb{Z}$. Also $e=b^d =(a^k)^d =a^{kd} $ and $|a|=n $ implies $n|kd$.

Hence $kd=nk'$ for some $k'\in \mathbb{Z}$ and $k=(\frac{n}{d})k'$

Hence $b=a^k =a^{{(\frac{n}{d})}k'}$ for some $k' \in \mathbb{Z}$ implies $b\in H_d$. Hence $H\subseteq H_d$ and then $H=H_d$.

Note: $|G|$ : order of the group $G$.

Is the proof correct ? Is there any mistake?

Answer 1

Here is my proof:

Existence:

Let $k=\frac{n}{d}\Longleftrightarrow k\cdot d=n$, so, $H=\langle a^k \rangle=\{e,a^k,a^{2k},...,a^{(d-1)k} \}$ and $|H|=d$,

Uniqueness:

Suppose $H_2$ is another subgroup of $G$ with order $d$, $|H_2|=d$.

Let $H_2=\langle a^q\rangle$, where $0\le q\le n-1$,

Case $(1)$:

If the generator $a^q\in H$,

then for any $x\in H_2, ~x=(a^q)^m$ for some integer $m$.

Since $a^q\in H$, $\Rightarrow (a^q)^m\in H, \Rightarrow x\in H\Rightarrow H_2\subseteq H$.

We also have $|H|=|H_2|=d$, therefore, $H_2=H$ .

Case $(2)$:

If the generator $a^q\notin H$, which means $k\nmid q~~~~~(*)$

Since $\frac{n}{\gcd(k,n)}=\frac{n}{k}=|H|=d=|H_2|=\frac{n}{\gcd(q,n)}$,

$\Rightarrow \gcd(k,n)=k=\gcd(q,n)\Rightarrow k\mid q~$, which contradicts with $(*)$.

Therefore, Case $(2)$ can never happen.

Proof is completed.

Answer 2

The proof seems good. Here's a different one. Consider the group homomorphism $\varphi\colon\mathbb{Z}\to G$, $\varphi(k)=a^k$.

This homomorphism is surjective and, by the homomorphism theorems, there is a bijection between the subgroups of $G$ and the subgroups of $\mathbb{Z}$ containing $\ker\varphi=n\mathbb{Z}$. This bijection is $H\mapsto\varphi^\gets(H)$ (inverse image), with inverse $N\mapsto \varphi^\to(N)$ (direct image). Moreover, given a (normal) subgroup $H$ of $G$, $\varphi$ induces an isomorphism $$ \varphi_H\colon\mathbb{Z}/\varphi^\gets(H)\to G/H $$ by $k+\varphi^\gets(H)\mapsto \varphi(k)H$.

The subgroups of $\mathbb{Z}$ containing $n\mathbb{Z}$ are of the form $d\mathbb{Z}$, for $d$ a divisor of $n$.

If $H$ has to be a subgroup of order $d$, we need $|G/H|=n/d$, so $\varphi^\gets(H)=(n/d)\mathbb{Z}$ by counting orders. This settles both existence and uniqueness.

Well, this is pretty abstract, but we can make it concrete. We have seen that the unique order $d$ subgroup has to be $\varphi^\to((n/d)\mathbb{Z})$ and the generator of such (cyclic) subgroup is $\varphi((n/d))=a^{n/d}$.

Answer 3

I have the same proof for uniqueness, but there is a formula that instantly shows exists.

Let $\frac{n}{k}=p$. We know that $a^p$ is in $G$ since $p<n$, thus we know that we can form a cyclic subgroup, $\langle a^p\rangle$. Now we just need to find the order of this subgroup, but we can use:

$$o(g^x)=\frac{o(G)}{\gcd{(o(G),x)}}$$ Here $x=p$ and $o(G)=n$ so, $$o(g^p)=\frac{n}{\gcd{(n,p)}}$$

Since $\frac{n}{k}=p$, we know $\gcd(n,p)=p$ thus $o(g^p)=\frac{n}{p}=d$.