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 $4$.
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.
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 $(xy)^2$.
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 $|F/H|\geq |D_4|$
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 $0\leq k<2$, $0\leq l<4$.
It follows from 1 and 2 that $F/H\cong D_4$.