= Solution
A <subgroup> $N\leq G$ is <normal subgroup>[normal] when
$$
gNg^{-1}=N\qquad(g\in G).
$$
Its left and right cosets then agree, and multiplication
$$
(gN)(hN)=ghN
$$
is well defined on the cosets. The identity is $N$, the inverse of $gN$ is $g^{-1}N$, and associativity descends from $G$, so these cosets form the <quotient group> $G/N$.
For a <group homomorphism> $\theta:G\to H$, its <kernel of a group homomorphism>[kernel] and <image of a group homomorphism>[image] are
$$
\ker\theta=\{g:\theta(g)=e_H\},
\qquad
\operatorname{im}\theta=\{\theta(g):g\in G\}.
$$
The kernel is normal because
$$
\theta(gkg^{-1})
=\theta(g)\theta(k)\theta(g)^{-1}
=e_H
$$
whenever $k\in\ker\theta$. Conversely, if $K\trianglelefteq G$, the quotient map
$$
q:G\longrightarrow G/K,\qquad q(g)=gK,
$$
is a homomorphism with kernel $K$. The image of any homomorphism is closed under products and inverses, so it is a subgroup of $H$. Finally, the map
$$
G/\ker\theta\longrightarrow\operatorname{im}\theta,
\qquad
g\ker\theta\longmapsto\theta(g)
$$
is well defined and bijective and preserves multiplication. This is the <first isomorphism theorem>.
Define
$$
\Phi:(\mathbb R,+)\longrightarrow(\mathbb C\setminus\{0\},\cdot),
\qquad
\Phi(t)=e^{2\pi it}.
$$
<Euler's formula> shows that $\Phi$ is a homomorphism, its image is the <complex unit circle>, and its kernel is $\mathbb Z$. The first isomorphism theorem therefore gives
$$
\boxed{\mathbb R/\mathbb Z\cong S^1}.
$$
The image of $\mathbb Q/\mathbb Z$ consists exactly of the <roots of unity>: if $t=p/q$, then $\Phi(t)^q=1$, while every element of finite order on the unit circle has an argument that is a rational multiple of $2\pi$.
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