Why is the binary dihedral group $BD_4$ the same as $\mathbb{Z}_4$? I am trying to understand why the binary dihedral group $BD_{4m}, m \in \mathbb{Z}$, with presentation $\langle\ A, B \mid\ A^{2m} =1,\ A^m = B^2 = -1,\ BAB^{-1}=A \rangle$ is the same, when $m=1$, as the additive group $\mathbb{Z}_4$. In the presentation of $BD_{4m}$, $A$ is
\begin{bmatrix}\epsilon&0\\0&\epsilon^{-1}\end{bmatrix}
and $B$ is 
\begin{bmatrix}0&-1\\1&0\end{bmatrix}
As far as I can see, when $m=1, BD_4 = \{e, A, B, AB\}$ where $e$ is the identity matrix, and we have that $\mathbb{Z}_4 = \{0,1,2,3\}$ and these are not very similar apart from both having four elements.
Any insight into why these two groups are the same would be appreciated!

I'm not sure what exactly is meant by $-1$, but I'm assuming its an element of your group not equal to the identity (well, I guess it is $A$?). In that case $B$ is not the identity and does not have order $2$ (because $B^2 \neq 1$), hence $B$ most have order $4$ by Lagranges theorem. But then your group is generated by $B$, i.e. is cyclic of order $4$ (for an explicit isomorphism try sending $B$ to $1 \in \mathbb Z_4$ since $1$ has order $4$ in that group).
