Can someone tell me what group this is? I ran across a group with presentation $<a,b | a^2b=ba^2, b^2a=ab^2, a^2=b^4>$.  Does this group have a name?
 A: Letting $W$ be an arbitrary word, using the third relation $a^2 = b^4$, we can assume that $W$ contains alternating $a$ and $b^k$ factors. Using the second relations $ab^2 = b^2a$ you can transform $W$ to the form 
$$ W = b^{k} (ab)^l \mbox{ or } b^k (ba)^l $$
But now note $(ba)(ab) = b^6 \implies ba = b^6(ab)^{-1}$, so every word $W$ is equivalent to 
$$ W = b^k (ab)^l $$
for some pair of integer $k,l$. Let us write out the multiplication table for this group in the presentation $(k,l)$. First, the identity is $(0,0)$. Observe that $(2k,0)$ commutes with all other elements. And $(k,l) = (k,0)(0,l)$, but $(0,l)(1,0) = (1+6l,-l)$. In other words, you can reduce to the group $\langle u,b | u^kb = b^{1+6k}u^{-k}\rangle$. 
Or, you can think of the group as the free abelian group $\mathbb{Z}^2 = \langle u,v | uv = vu\rangle$ plus the additional generator $b$ where $b^2 = v$ and $b^{-1}ub = b^6 u^{-1}$. (Nope, I also don't know what this group is, but maybe someone else can take it from here.)
A: Let $G = \langle a, b\ |\ [a,b^2]=1, a^2=b^4\rangle$ be your group.  Consider the map of $G$ to $\mathbb{Z}$, sending $b$ to $1$ and $a$ to $2$.  This has a kernel which is $D_\infty$ (found via Reidemeister-Schreier rewriting).  That is, the kernel is $\langle x,y\ |\ x^2=y^2=1\rangle$, where $x=ab^{-2}$ and $y=bab^{-3}$.  Thus your group splits as a semidirect product $D_\infty \rtimes\mathbb{Z}$, where if $D_\infty=\langle x,y\ |\ x^2=y^2=1\rangle$ and $\mathbb{Z}=\langle t\rangle$, then the action of $t$ is: $x^t=y$, $y^t=x$.
EDIT: So we've gotten the following presentation of your group:
$$ G= \langle x,y,t\ |\ x^2=y^2=1, x^t=y, y^t=x\rangle.$$
Let $K=\langle xy,t\rangle$ be a subgroup of $G$; it is not hard to see $K$ has index 2, and is isomorphic to the Klein bottle group.  Thus your group $G$ is also a semidirect product $K\rtimes C_2$ (here $C_2$ is the cyclic group of order two), where if $C_2=\langle u\rangle$, and $K=\langle r,s\ |\ s^{-1}rs=r^{-1}\rangle$, then $r^u=r^{-1}$ and $s^u=sr^{-1}$.
