Is it true that every mapping class in $\mathrm{Mod}(\Sigma_3)$ commutes with some hyperelliptic involution? Two questions.
First, let $\Sigma_3$ be the closed genus 3 surface and let $\rm Mod(\Sigma_3)$ be its mapping class group. Is it true that for any mapping class $g\in\rm Mod(\Sigma_3)$ there is some hyperelliptic involution of $\Sigma_3$ which it commutes with? By hyperelliptic involution, I mean a mapping class with an order two representative $f$ with $8$ fixed points.
Also, another question, are there any results on choosing representatives for each coset of $\rm HMod(\Sigma_3)$, the hyperelliptic mapping class group?
Thank you for any help.
 A: No. After the Nielsen realization theorem, this follows from the classification of automorphism groups of genus three Riemann surfaces, which can be found in many places. For example, the cyclic group $C_9$ acts on the curve with projective equation $x^4 + xy^3 + yz^3$. But there is no hyperelliptic Riemann surface of genus three with an automorphism of order $9$.
A: V.I. Arnold proved that the image of the hyperelliptic group in the mod 2 symplectic group (ie, the image of $HMod_g\to Mod_g\to Sp_{2g}(\mathbb Z)\to Sp_{2g}(\mathbb F_2)$) is the symmetric group on the branch points: $HMod_g\twoheadrightarrow Mod_{0,2g+2}\twoheadrightarrow S_{2g+2}\hookrightarrow Sp_{2g}(\mathbb F_2)$. For $g=1,2$, this inclusion of finite groups is an isomorphism, but not for $g\geqslant 3$. Moreover, for $g=3$, not every element is conjugate into the symmetric group. In particular, there are elements of order 9 in $Sp_6(\mathbb F_2)$ but not in $S_8$. This is an exercise in linear algebra which I messed up when preparing this answer, so me rely on the symmetry Dan mentioned, which is thus a finite element of the mapping class group, and thus retains its order in the coprime symplectic group. Thus not only does this finite element of the mapping class group not commute with any hyperelliptic element, no element which reduces to it in $Sp_6(\mathbb F_2)$ commutes with any hyperelliptic element.
(Linear algebra: given a Galois extensions $K\subset M$ which is a composite of two Galois extensions $M=L_1L_2$ of degrees $g$ ($L_1/K$) and $2$ ($L_2/K$); put a symplectic form on $M/K$ so that $U_1(M/L_1)\subset U_g(L_2/K)\subset Sp_{2g}(K)$. Ultimately $K=\mathbb F_2$, $L_1=\mathbb F_8$, and $L_2=\mathbb F_4$; so $U_1(F_{64}/F_{8})=\mu_9\subset Sp_6(\mathbb F_2)$.)
A: An observation relating to your second question:
Fix a hyperelliptic involution $\iota$, and let $c$ be a (unoriented isotopy class of a) curve not fixed by $\iota$.  Denote a (right) Dehn twist about $c$ by $T_c$.
The set $\{T_c^n:n \in \mathbb Z\}$ is a set of distinct coset representatives of $\operatorname{HMod}(\Sigma_3)$ in $\operatorname{Mod}(\Sigma_3)$ (since $\operatorname{HMod}(\Sigma_3)$ is the centralizer of $\iota$ in $\operatorname{Mod}(\Sigma_3)$, and $\iota T_c^n \iota^{-1} = T_{\iota(c)}^n$).  This provides infinitely many coset representatives, but at first glance there is no reason to believe that this is a complete set of coset representatives.
Other than this observation, I am unaware of any work about choosing coset representatives of the hyperelliptic mapping class group, but that's not to say they're not out there!
