Why is the binary dihedral group $BD_4$ the same as $\mathbb{Z}_4$?

Question

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!

Answer 1

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).