I wrote up a general classification for the centers of $D_n$, (the dihedral group of order $2n$, not $n$) just the other week. Perhaps it will be useful to read:
If $n=1,2$, then $D_n$ is of order $2$ or $4$, hence abelian, and $Z(D_n)=D_n$. Suppose $n\geq 3$. We have the presentation
$$
D_n=\langle x,y:x^2=y^n=1,\; xyx=y^{-1}\rangle.
$$
Then $yx=xy^{-1}$ implies the reduction $y^kx=xy^{-k}$. An element is in the center iff it commutes with $x$ and $y$, since $x$ and $y$ generate $D_n$. Let $z=x^iy^j$ be in the center. From $zy=yz$ we see
$$
x^iy^{j+1}=yx^iy^j\implies x^iy=yx^i.
$$
But $i\neq 1$, else we have $xy=yx=xy^{-1}$, so $y^2=1$, a contradiction since $n\geq 3$. So $i=0$, and $z=y^j$. Then from the equation $zx=xz$, we have
$$
y^jx=xy^j=xy^{-j}
$$
which implies $y^{2j}=1$. Thus $j=0$ or $j=n/2$. If $n$ is odd, we must necessarily have $j=0$, and $z=1$. If $n$ is even, either possibility works. But $y^{n/2}$ is indeed in the center as it clearly commutes with $y$, as well as with $x$ since $y^{n/2}x=xy^{-n/2}=x(y^{n/2})^{-1}=xy^{n/2}$. Summarizing, we have, for $n\geq 3$,
$$
Z(D_n)=\begin{cases}
\{1,y^{n/2}\} & \text{if }n\equiv 0\pmod{2},\\
\{1\} & \text{if }n\equiv 1\pmod{2}.
\end{cases}
$$
First, let's work purely algebraically:
The elements of $\mathbb{Q} / \mathbb{Z}$ are the cosets $q + \mathbb{Z}$ for each rational $q \in \mathbb{Q}$. So $q_1$ and $q_2$ become equal if they differ by an integer. So we are considering $\frac{1}{2}$ and $\frac{3}{2}$, for instance to be the same. So for any $q \in [n, n+1)$ we can shift it to be in $[0,1)$ by subtracting $n \in \mathbb{Z}$. So we see that $\mathbb{Q} / \mathbb{Z}$ is exactly like $\mathbb{Q} \cap [0,1)$, where we work "mod 1". When we add $\frac{2}{3} + \frac{1}{2}$, we get $\frac{7}{6}$, which we reduce to $\frac{1}{6}$.
Geometrically, this is like "winding $\mathbb{Q}$ around a circle". It is at this point that I am legally required to show you this picture:
We are wrapping $\mathbb{Q}$ into a circle, so that two points in $\mathbb{Q}$ lie on top of each other exactly when their difference is an integer.
When you internalize these two perspectives, you should see why every element of $\mathbb{Q} / \mathbb{Z}$ has finite order. After all, if we start with $\frac{a}{b}$, and we add it to itself $b$ times, we'll be left with
$$
\underbrace{\frac{a}{b} + \frac{a}{b} + \cdots + \frac{a}{b}}_{b \text{ times}}
= b \frac{a}{b} = a = 0 \text{ (mod } 1)
$$
Now let's move to $\mathbb{R} / \mathbb{Q}$. This one is extra tricky, because there is no constructive way to choose an element from each coset. (There is some set-theoretic subtletly regarding the axiom of choice
Much ink has been spilled trying to understand this group (see
here and
here just to start), but I will give one "elementary" way to understand what is happening (though trying to visualize this group is going to be a mess no matter what you do). At the very least, this will let us resolve the question about infinite order elements.
We can write $\mathbb{R}$ as a Vector Space over $\mathbb{Q}$, and so (assuming choice) we can find a basis for $\mathbb{R}$ as a vector space. Just like every finite dimensional vector space over $\mathbb{R}$ looks like $\mathbb{R}^n$ for its dimension $n$, every infinite dimensional vector space over $\mathbb{Q}$ looks like $\mathbb{Q}^\kappa$ for its dimension $\kappa$.
So $\mathbb{R} \cong \mathbb{Q}^{\mathfrak{c}}$ (here $\mathfrak{c} = |\mathbb{R}| = \text{dim}(\mathbb{R}$) over $\mathbb{Q}$). Then when we quotient out by $\mathbb{Q}$, we can think about this as killing off one of the $\mathfrak{c}$ copies of $\mathbb{Q}$. Of course, since $\mathfrak{c}$ is infinite, we have (set theorists look the other way) "$\mathfrak{c} - 1 = \mathfrak{c}$". So we see that:
$$
\mathbb{R} / \mathbb{Q} \cong
\mathbb{Q}^{\mathfrak{c}} / \mathbb{Q} \cong
\mathbb{Q}^{\mathfrak{c}} \cong
\mathbb{R}
$$
Thus, by some nonconstructive black magic, $\mathbb{R} / \mathbb{Q} \cong \mathbb{R}$. But there is no way to describe the isomorphism more than we already have.
Like I said, though, this is enough information to answer your question:
Since every nonzero element of $\mathbb{R}$ has infinite order, so too does
every nonzero element of $\mathbb{R}/\mathbb{Q}$.
I hope this helps ^_^
Best Answer
Note that centre of $D_6$ has two elements, namely, identity and $R_{180}$. So order of $R_{60}Z(D_6)$ is three not six
Also the order of the quotient group is six. So the other way to see your calculation is wrong is by considering quotient group. If it we're cyclic then G is Abelian. But $D_6$ is non Abelian