The Problem: Show that $D_{2^n}/Z(D_{2^n})\cong D_{2^{n-1}}$.
I know that $Z(D_{2^n})=\{1, r^{2^{n-1}}\}$, so I figured that establishing an explicit isomorphism $\phi: D_{2^n}/Z(D_{2^n})\to D_{2^{n-1}}$ would be the way to go; but I failed to come up with a valid function. It seems like a really silly place to get stuck-any help would be greatly appreciated.
Remember that a group isomorphism is determined by where it sends the generators. If we say that $D_{2^{n-1}} = \langle a , b \; | \; a^{2^{n-1}} = b^2 = 1, \; ab = ba^{2^{n-1}-1}\rangle$, then you need to specify where in $D_{2^n}/Z(D_{2^n})$ both generators are to be mapped to. Why not define $\phi: D_{2^{n-1}} \to D_{2^n}/Z(D_{2^n})$ by mapping $\phi(a) = aZ(D_{2^n})$ and $\phi(b) = bZ(D_{2^n})$? Are you able to verify that this is the desired isomorphism?
The big takeaway here is that in many cases where you are asked to determine an explicit isomorphism, there is usually a very natural way of doing so, just like the one above.