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Studying for an exam, a review question...

Given $G=\langle x,y|x^4=y^4=e,xyxy^{-1}=e\rangle$.

  1. Show $|G|\leq16$.

For this, I want to consider that $x^3=x^{-1}$ and $y^3=y^{-1}$ based on our assumptions. I am a little lost as to how to put the second part, $xyxy^{-1}=e$ to show there are no more than 16 elements.

  1. If $|G|=16$, find the center of the group and find a group that is isomorphic to $G/\langle y^2\rangle$.

I'm pretty sure the group is centerless.

allthatmath
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    For the first part, note that the relation allows you to exchange $x$ and $y$ at the cost of adding an exponent. For the second part, it cannot be centerless if it has order $16$, as then it is a $2$-group. – Tobias Kildetoft Jun 25 '15 at 20:57

2 Answers2

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For the first part note that $yx=x^3y$ from the relations, so any element can be written as $x^ay^b$

For the second part all the forms $x^ay^b$ are distinct - otherwise the number of elements falls below $16$. If you have no better idea show that $x^ay^bx^cy^d=x^{a+3^bc}y^{b+d}$ and identify elements of the centre from this.

That should help you to complete things.

Mark Bennet
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First show that all elements of $G$ can be written as $x^iy^j$ with $0\leq i \leq 3$ and $0\leq j \leq 3$. This tells you that $|G|\leq 16$. Then find a semi-direct product of two cyclic groups of order $4$ that is the image of your group G. This shows that $|G|\geq 16$.

Chuks
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