Class equation, center and commutator subgroup I'm trying to determine the class equation, center and commutator subgroup of the group $G=\langle x, y\mid x^4,y^4,yxy^{-1}x\rangle$ 
First of all, I tried to determine $|G|$ directly. I could figure out $|y|=4$, but the rest didn't come... By observation I'm guessing $|x|=2$?
Any help greatly appreciated.
 A: Here, I am writing some achieved points and I hope these be useful for solveing the problem completely. Let's find the presentation of the quotient group $G/G'$: $$G/G'=\langle x,y\mid x^4=1,y^4=1,yxy^{-1}x=1,[x,y]=1\rangle$$ which is
$$G/G'=\langle x,y\mid x^4=1,y^4=1,x^2=1,[x,y]=1\rangle\\\ \cong\langle x,y\mid x^2=1,y^4=1,[x,y]=1\rangle\cong\mathbb Z_2\times\mathbb Z_4$$ Moreover if $H=\langle x\rangle$ then $|H|=4$ and as you and @Steve D remarked it is normal in the group $G$.  Now let's find the presentation of $G/H$:
$$G/H=\langle x,y\mid x^4=1,y^4=1,yxy^{-1}x=1,x=1\rangle\cong\langle x,y\mid x=1,y^4=1\rangle\cong\mathbb Z_4$$ so $G/H$ is abelian and so $G'\subseteq H$. Then:


*

*$G'=\{e\}$ and so $G\cong\mathbb Z_2\times\mathbb Z_4$ which is a contradiction.

*$|G'|=2$ then $G'\cong\langle x^2\rangle$

*$|G'|=4$ then $G'\cong\langle x\rangle$

A: I see that Babak Sorouh has already answered, but I propose a different solution.
I want to prove that $G=\mathbb Z_4 \rtimes_\varphi \mathbb Z_4$ where $\varphi \colon \mathbb Z_4 \to \text{Aut}(\mathbb Z_4)$ is the homomorphism sending the generator of $\mathbb Z_4$ in the inverse homomorphism (i.e. the map $x \to -x$).
First of all remember that a presentation denote a group which is the biggest quotient of the free group on the generators, meaning that for every every group $\tilde G$ having two generators which satisfy the relations there's a unique homomorphism $G \to \tilde G$ which preserve generators.
Clearly $\mathbb Z_4 \rtimes_\varphi \mathbb Z_4$ satisfies the relations: indeed is generated by the elements $(1,0)=x$ and $(0,1)=y$, which have both order $4$ and $(0,1)(1,0)(0,-1)=(-1,0)$ which is exactly the relation $yxy^{-1}=x^{-1} \iff yxy^{-1}x=\text{id}$. 
Given every other group $\tilde G$ generate by two elements $x,y \in \tilde G$ such that $x^4=y^4=\text{id}$ and $yxy^{-1}=x^{-1}$ in this group we have:


*

*two subgroups $H=\langle x \rangle$ and $K=\langle y \rangle$;

*$H$ is normal by the last of the relations, and so $$HK=\{ hk| h \in H,\ k \in K\}=KH=\{kh|k \in K,\ h \in H\}$$

*from what just said it follows that $\tilde G = HK = \{x^iy^j|i,j=0,\dots,3\}$.
So we can consider the map $f \colon \mathbb Z_4 \rtimes_\varphi \mathbb Z_4 \to \tilde G$ which sends every pair $(n,m)$ in $f(n,m)=x^ny^m$, which is surjective. A simple calculation (using the relations) shows that 
$$f((n,m)(n',m'))=f(n+(-1)^m n',m+m') = x^{n+(-1)^mn'}y^{m+m'} = x^ny^{m}x^{(-1)^{2m}n'}y^{-m}y^{m+m'}$$
which become
$$x^ny^mx^{n'}y^{m'}=f(n,m)f(n',m')$$
So $f$ is an homomorphism, and clearly is the unique such that $f(1,0)=x$ and $f(0,1)=y$.
So $G$ is indeed $\mathbb Z_4 \rtimes_\varphi \mathbb Z_4$ at this point I hope you'll find easy to calculate the center, the commutator subgroup and the class equation. 
Anyway if there's any trouble feel free to ask.
