Where $a$ = $(1 4 6 3 7)(2 5 3 8)(4 7 8 1) = (1 4 7)(2 5)(3 8)(6)$ as a product of disjoint cycles.
Is there an easier way than to compute every possible $b$ ? Like a conjecture for the form of $b$ ?
Where $a$ = $(1 4 6 3 7)(2 5 3 8)(4 7 8 1) = (1 4 7)(2 5)(3 8)(6)$ as a product of disjoint cycles.
Is there an easier way than to compute every possible $b$ ? Like a conjecture for the form of $b$ ?
We are looking for elements which commute with $a$, in other words we try to determine the centralizer $C_G(a)$, where $G=\mathbf S_8$. Here is an outline of what you can do:
Step 1: the following elements will certainly commute with $a$:
Now observe that $\langle s,x\rangle \cong\mathbf D_8$ is a dihedral group of order $8$, since it it generated by $2$ involutions whose product has order $4$. Together with the $\mathbf C_3$ this gives us a group $\mathbf C_3\times\mathbf D_8\leq C_G(a)$.
Step 2: by the orbit-stabilizer formula, $[G:C_G(a)]=|a^G|$. This orbit has $$ M = \binom{8}{3}2 \cdot 5 \cdot 3 $$ elements. (Pick $3$ random elements, order them in one of two ways; pick the last element; then split the remaining $4$ elements in half.)
But this says precisely that $|C_G(a)|=8!/M = 24$.
Conclusion: the elements above will precisely generate $C_G(a)$.
Post scriptum: I'm somewhat convinced that this generalizes to more complicated cycle decompositions, and that in general the centralizer of an element can be written as a direct product $$ \prod_i (\mathbf C_{m_i} \wr \mathbf S_{n_i}), $$ where there are $n_i$ cycles of length $m_i$. (Here we have the special case $(\mathbf C_3\wr\mathbf S_1) \times (\mathbf C_2\wr\mathbf S_2) \times (\mathbf C_1\wr\mathbf S_1)$). But proving it requires doing the above count in greater generality.