# 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.

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No, $|x|=4$, and in fact $\langle x\rangle$ is normal (can you see why?). The quotient $G/\langle x\rangle$ is generated by $y$, so the commutator is contained in $\langle x\rangle$. The last relation of your presentation gives a very good idea what the center is. – user641 Jan 15 '13 at 8:52
It is quite late at night... Would you mind elaborating some more? (and writing it as an answer so that I could click accept for you?) – Benjamin Lu Jan 15 '13 at 8:56
I would actually prefer you showed a bit more work, before I decided to write an answer. But there is no rush; get some sleep, and see what you can accomplish tomorrow. :) – user641 Jan 15 '13 at 8:57
I can see how $<x>$ is normal because according to the last relation conjugation by any element would send elements of $<x>$ back to $<x>$ itself. x is of order 1, 2, or 4 I suppose. Why can't it be 1? – Benjamin Lu Jan 15 '13 at 9:06
Do you know about semidirect products? This group is a semidirect product of two cyclic groups of order 4. – Derek Holt Jan 15 '13 at 9:48

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 $G'\cong\langle x\rangle$
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How did this go unnoticed (or "un-up-voted") for so long! + 1 – amWhy Feb 16 '13 at 0:16
@amWhy: Thanks for your kind deeds. Sometimes I made some answers here, but didn't get any responses. Anyway, I love Maths just for Maths. Love your responses. – Babak S. Feb 16 '13 at 6:18

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.

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