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I want to find the presentation of group $D_4= \{1, f, r,r^2,r^3, rf, r^2f, r^3f \} $. $r$ is the rotation of a square counterclockwise by 90 degree and $f$ is the action that flips the square. Here $f$ has order $2$ and $r$ has order $4$. enter image description here

I have found that the relations $$f^2 =1 \\ r^4 =1 \\ fr=r^3f$$ can deduce all the operation results of elements in $D_4$. However, I am not sure $D_4 =\langle f,r | f^2 =1 ,r^4 =1 , fr=r^3f \rangle $ is the presentation of $D_4$. In other words, is $D_4$ the biggest group having these relations?

I feel that equality is key to this question. One must show there are no other equalities of two elements except those deduced from the abovementioned generating relations.

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    $\begingroup$ See Theorem 1.1 in kconrad.math.uconn.edu/blurbs/grouptheory/dihedral2.pdf for this result for all dihedral groups $D_n$ when $n \geq 3$. $\endgroup$
    – KCd
    Jun 13, 2022 at 13:49
  • $\begingroup$ I vote to reopen the question. This question is not a duplicate of the question Why must a group with presentation The question at the link and the first answer to it has to do with defining the concept of generators and relations on these generators. The question under discussion is about computing the presentation of a particular group. $\endgroup$
    – kabenyuk
    Jun 14, 2022 at 5:46

1 Answer 1

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The usual reasoning is this.

Hints.

Let $F=\operatorname{gr}(x,y)$ be a free group with free generators $x$ and $y$, and $H$ be a normal subgroup generated by $x^2$, $y^4$, and $(xy)^2$.

  1. Consider the homomorphism $\phi:F\to D_4$ defined by the formulas $\phi(x)=f$, $f(y)=r$. Clearly, $H\leq\operatorname{Ker}(\phi)$ and $\operatorname{Im}(\phi)=D_4$. Therefore the factor-group $|F/H|\geq |D_4|$

  2. Let us prove that $|F/H|\leq8$. To do this, work with words in the group $F$. Using the fact that $x^2,y^4,(xy)^2\in H$ we prove that $$ x^{k_1}y^{l_1}\ldots x^{k_s}y^{l_s}H=x^ky^lH, $$ where $k_i,l_i,k,l\in\mathbb{Z}$ and $0\leq k<2$, $0\leq l<4$.

  3. It follows from 1 and 2 that $F/H\cong D_4$.

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  • $\begingroup$ Thanks for your answer! The concept of free group and quotient group is new to me. But this is very helpful. Is it also true that $F/H$ is isomorphic to any group of order 8 with the relation $x^2=x^4=(xy)^2=1$, so that the biggest group with these relations is unique? $\endgroup$
    – user628623
    Jun 14, 2022 at 5:05
  • $\begingroup$ btw should the notations in (2.) be $k<2$ and $l<4$ instead of $\leq $? $\endgroup$
    – user628623
    Jun 14, 2022 at 5:08

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