abstract algebra - Proving the factor group is isomorphic to $mathbb{Z}_8$ .

Let $G= U(32)$ and $K$= {${1,15}$}.
Prove that $G/K$ is isomorphic to $\mathbb{Z}_8$.

I know that $|G/K|$= $|U(32)| / |K|$ = $16/2 =8$.

$G/K =$ {$ K, 3K, 5K, 7K, 17K, 19K, 21K, 23K$}.

While trying to see that the properties of isomorphic groups work on $G/K$ and $\mathbb{Z}_8$, I tried to find the number of elements of order $8$ in $G/K$. I found that $3K$, $5K$, $19K$ and $21K$ have order $8$ in $G/K$ like $1, 3, 5$ and $7$ in $\mathbb{Z}_8$.

I know that this can't prove isomorphism but really I can't approach the solution. Maybe the Cayley Table may help but not in PROVING I think.

Frankly, I'm not strong enough at isomorphism or normal groups as my professor illustrated them quickly in like 2 lectures, and I have a final exam.

Answer

$\newcommand{\Int}{\mathbf{Z}}$Let $n$ be a positive integer. A cyclic group $(G, \cdot)$ of order $n$ is isomorphic to the "standard" cyclic group $(\Int_{n}, +)$: If $a \in G$ is a generator, then every element of $G$ may be written in the form $a^{r}$ for a unique integer $r$ with $0 \leq r < n$, and the group law is
$$
a^{r} \cdot a^{s} = a^{r + s} = a^{r + s \pmod{n}}
$$

by the law of exponents and the fact that $a^{n} = e$ in $(G, \cdot)$.

In fancy language, the mapping $\phi:\Int_{n} \to G$ defined by $\phi(r) = a^{r}$ induces an isomorphism of groups $(\Int_{n}, +) \to (G, \cdot)$ because
$$
\phi(r + s) = a^{r + s} = a^{r} \cdot a^{s} = \phi(r) \cdot \phi(s)
$$
for all integers $r$ and $s$.

In your example,
\begin{alignat*}{2}

(3K)^{1} &= \{ 3, 13\} &&= 3K, \\
(3K)^{2} &= \{ 9, 7\} &&= 7K, \\
(3K)^{3} &= \{27, 21\} &&= 21K, \\
(3K)^{4} &= \{17, 31\} &&= 17K, \\
(3K)^{5} &= \{19, 29\} &&= 19K, \\
(3K)^{6} &= \{23, 25\} &&= 23K, \\
(3K)^{7} &= \{ 5, 11\} &&= 5K, \\
(3K)^{8} &= \{15, 1\} &&= K.
\end{alignat*}
This shows the mapping $\phi:\Int_{8} \to U(32)/K$ defined by $\phi(r) = (3K)^{r}$ induces an isomorphism. (I'm being picky by saying "induces" rather than "is": The mapping acts on sets, while isomorphism is a property of groups—sets equipped with group operations.)

The other generators you found induce distinct isomorphisms, four in all.