Find all $x$ in $D_4$ such that $ax(a^{-1}) = b$.

Question

Consider the dihedral group $D_4$. Consider also the elements $a= r_1$ and $b= S_1$ of $D_4$. Find all $x$ in $D_4$ such that $ax(a^{-1}) = b$. Do both $a$ and ($a^{-1}$) cancel each other out? If not, how do I solve this?

Answer 1

$D_n := \langle r, s : |r| = n, |s| = 1, rs = sr^{-1}, s \neq r^i, \forall i \rangle, n \geq 3$ be the dihedral group of order $2n$ with the standard generators ($r$ is the rotation by $\frac{2 \pi}{n}$ and $s$ is reflection through the vertex $1$ and origin). Then every element of $D_n$ can be written uniquely as $r^is^j, 1\leq i \leq n, j = 1, 2.$

In this case $n = 4$ and with this notations $r_1 = r, s_1 = s$ (if I understand correctly). So we need to find all $x \in D_4$ such that $rxr^{-1} = s.$ Now $x$ is of the form $r^is^j.$ First let $j = 0.$ Then $rxr^{-1} = rr^ir^{-1} = r^i.$ So there is no solution in this case. Now let $j = 1.$ Then $rxr^{-1} = rr^isr^{-1} = r^{i+2}s.$ Hence $rxr^{-1} = s \Rightarrow r^{i+2}s = s \Rightarrow r^{i+2} = 1 \Rightarrow i = 2.$ In this case, the only choice for $x$ is $r^2s.$

Obviously, you can do it more directly as follows: $rxr^{-1}=s \Rightarrow x = r^2s.$