Show that the dihedral group $ D_3 $ has only one subgroup of order 3 The extra hint was that the order $ \circ (g) $ of $g$ divides the order of $ \circ (G) $ of $G$
I claimed that the subgroup of order $3$ in $G$ was $\langle\,a\,\rangle_{a^3=e}$ which was obvious. How do I prove that this is the ONLY subgroup?
 A: 
In $D_3$, we have $3$ rotations ($e, R, R^2)$and $3$ reflections (each reflection is of order $2$) so subgroup of order $3$ must contain elements only from $e, R, R^2$. Clearly, it's only one subgroup of order $3$, which is cyclic i.e. $\langle R \rangle $. 
On the contrary, suppose that there is another subgroup of order $3$. Let's call it $\langle b \rangle $ such that $b\notin \langle R \rangle $. This implies that $|<R>\cap<b>|=1$.
$|\langle R \rangle \langle b \rangle|=\frac{|<b>||<R>|}{|<R>\cap <b>|}=9\gt |D_3| $, which is a contradiction. 
As per your comment, since the above theorem has not been taught yet in your class, You may refer to this part:
No. of order $3$ elements in a subgroup (of $D_3$) of order $3= \phi(3)=2$. No. of order 3 elements in $D_3$ are $R, R^2$ so there can be only one subgroup viz. $\{e, R, R^2\}$
A: We have to prove that $\langle r\rangle=\{1,r,r^2\}$ is the only subgroup of order $3$ of $D_3=\{1,r,r^2,s,rs,r^2s\}$. Call $H$ a candidate subgroup of order $3$ distinct from $\langle r\rangle$. Note that, by closure, both $r\in H \Rightarrow r^2\in H$ and $r^2\in H \Rightarrow r\in H$, and thence $H\cap\langle r\rangle \ne\{1\} \Rightarrow H=\langle r\rangle$. Therefore, necessarily $H\cap \langle r \rangle=\{1\}$. So, the only options left are:

*

*$\space H=\{1,s,rs\}$

*$\space H=\{1,s,r^2s\}$

*$\space H=\{1,rs,r^2s\}$
Options 1, 2 and 3 are discarded because, respectively:

*

*$\space rss=rs^2=r\notin H \Rightarrow H$ is not a subgroup;

*$\space r^2ss=r^2s^2=r^2 \notin H\Rightarrow H$ is not a subgroup;

*$\space rsr^2s\stackrel{(rs)^2=1}{=}sr^{-1}r^2s=srs=ssr^{-1}=r^2\notin H\Rightarrow H$ is not a subgroup.

Therefore, $\langle r \rangle$ is the only subgroup of $D_3$ of order $3$.

If you had at hands the result mentioned in Bungo's comment, than we could prove that this claim holds for every $D_p$, with $p$ odd prime. In fact, again in this case we have $H\cap \langle r\rangle=\{1\}^{(*)}$, and thence:
$$|H\langle r\rangle|=\frac{o(H)o(r)}{o(H\cap\langle r\rangle)}=\frac{p^2}{1}=p^2 \tag 1$$
But $p^2>2p=|D_p|$ as soon as $p>2$, so there isn't "enough room" in $D_{p>2}$ to accomodate such a $H$.
$^{(*)}$Otherwise:
$$\exists j\in \{1,\dots,p-1\} \mid r^j\in H\cap \langle r\rangle \Rightarrow r^j\in H\cap \langle r\rangle, \forall j\in\{1,\dots,p-1\}\Rightarrow H=\langle r\rangle$$
