For $ n \in \mathbb{N}$ and $0 \le r <n$, define $f_r : \mathbb{C} \to \mathbb{C}$ ; $z \mapsto ze^{2\pi i r/n}$ and $c: \mathbb{C} \to \mathbb{C}$ ; $z \mapsto \bar{z}$.
a) Let $D_n = \{ f_0, ..., f_{n-1}, f_0 \circ c, ..., f_{n-1} \circ c \}$. Show that $D_n$ is a subgroup of Perm($\mathbb{C}$), where Perm($\mathbb{C}$) is the set of all invertible self-maps on $\mathbb{C}$.
b) Show $D_4$ is not isomorphic to $Q_8$.
Here's what I have so far:
First I said that $f_k \circ c (z) = f_k(\bar{z}) = \bar{z}e^{2\pi i k/n} = ze^{2 \pi i (n-k)/n} = f_{n-k}$, since taking the conjugate just reflects $x$ across the real axis. So $f_k \circ c = f_{n-k}$ for any $k$.
To check closure, then, just consider two elements $f_k$ and $f_j$. Then $f_k \circ f_j = f_{k+j (mod n)}$. For inversion, $f_y^{-1} = f_{n-y}$, so all inverses are in $D_n$.
Hint: the relation between rotations and reflections looks like this: $$ (f_k \circ c) (z) = f_k(\bar{z}) = \bar{z}e^{2\pi i k/n} = c(ze^{2 \pi i (n-k)/n}) = (c \circ f_{n-k}) (z) $$
Now you should calculate each of the following expressions (for $0 \le k < n$) and use the relation to express the result in the form $f_j$ or $f_j \circ c$ for some $j$ (these are the elements of the group that you provided). You can use the relation at the level of functions $(f_k \circ c = c \circ f_{n - k})$, so you don't have to drag the argument $z$ through your calculation.
Closure under composition:
Closure under inverses: