If $G$ is finite simple and $H \le G$ has index $2m$, then an involution of $G$ is conjugate to one in $H$ I have been trying to prove the following for a while now, with plenty of attempts and ideas, but no success.  I would appreciate a gentle nudge in the right direction.
Suppose that $G$ is a finite simple group with $|G|>2$.
a) Let $H \le G$, where $|G:H|=2m$, and $m$ is odd.  If $t \in G$ has order $2$, show that $t$ is conjugate in $G$ to some element of $H$.
b) Suppose that a Sylow $2$-subgroup of $G$ is isomorphic to the dihedral group $D_8$ (of order $8$).  Show that $G$ has a unique conjugacy class of involutions.
For (a) I am given the hint to let $G$ act on the right cosets of $H$ via right multiplication.
Let $Ha$, $Hb$ be distinct right cosets of $H$ in $G$.  Then $Ha\cdot (a^{-1}b)=Haa^{-1}b=Hb$, so the action is transitive.  It follows that there is only one orbit, with size $2m$.  By the orbit-stabilizer theorem, this tells us that $|G:G_{Ha}|=2m$ for any $Ha \in G/H$ ($G_{Ha}$ is the stabilizer of $Ha$ in $G$).  I have been racking my brain trying to find a way to show that $t$ fixes some $Ha$, because then $Ha \cdot t = Hat = Ha$, and it follows that $Hata^{-1}=H$ and $ata^{-1} \in H$, giving the desired result.
I'm not sure how to use the assumptions that $G$ is simple and $|G:H|=2m$.  What I can say is that the kernel of the action of $G$ on $H$ must be trivial.  Already I mentioned that $|G:H|=2m$ gives us the size of the single orbit is $2m$, and this says that the $|G:G_{Ha}|=2m$ for each $Ha$.  I'm not sure what the significance of $m$ being odd is, because G/H is only a set and not a group (in this case).  I haven't managed to deduce any useful group properties from it, either.
Another possibility is to show that $|H|$ is divisible by $2$.  Then $H$ contains an involution which is contained in some Sylow $2$-subgroup of $G$, and all of these (there must be more than one since $G$ is simple) are conjugate, so I think this may give us what we need.
Again, I'd appreciate a hint (only) on how to proceed.  Thanks.
 A: For (b), I can use (a) to show that there are at most two conjugacy classes.
Let $P$ be a Sylow 2-subgroup, let $t$ be an involution distinct from the involution $z$ in $Z(P)$, and let $H=C_G(t)$ be the centralizer of $t$ in $G$. Since $t$ and $z$ are in $H$, we get that a Sylow 2-subgroup of $H$ is either of order 4 or 8.
Order 4:  [ In reality, this case can never occur. You prove this by transfer or fusion. ]

 If order 4, then $[G:H]=2m$, and so every involution is conjugate to either $t$, $tz$, or $z$, but $tz$ and $t$ are already conjugate in $P$.

Order 8:

 If order 8, then clearly $t$ is central in some other Sylow $P^g$, but since $z$ is the unique central involution of $P^g$, we get that $t=z^g$.

The other involutions are already handled:

 If $s$ is an arbitrary involution, then either $s$ is conjugate to $z$, $s=z^g$ and thus $s$ is central in $P^g$, or the centralizer $H=C_G(s)$ of $s$ has order not divisible by 8, and so  case 4 (strengthened by $s$ not conjugate to $z$) gives $s$ is conjugate to $t$.

In reality though, if the order 4 case ever happens you get that $G$ has a normal subgroup of index 2 by transfer.
A: I think I can continue Jack Schmidt's argument to complete the proof of (b) without  using transfer. A Sylow $2$-subgroup $S$ of $G$ is generated by two involutions $s,t$, where $z=(st)^2$ is the central involution.
If $s$ is conjugate in $G$ to $z$ then, since $s$ and $sz$ are conjugate in $S$, all three involutions in the $4$-group $T = \langle s,z \rangle$ are conjugate in $G$. Since $|G:T| = 2m$ with $m$ odd, $t$ (and $tz$) must also be conjugate to them, and so all involutions in $G$ are conjugate. The same applies if $t$ and $z$ are conjugate.
Since Jack has shown that there are at most two classes of involutions, the only other possibility is that $s$ and $t$ are conjugate in $G$, but are not conjugate to $z$. In that case, $\langle s,z \rangle$ and $\langle t,z \rangle$ must be Sylow $2$-subgroups of $C_G(s)$ and $C_G(t)$. If $s^g=t$, then $g$ conjugates a Sylow $2$-subgroup of $C_G(s)$ to one of $C_G(t)$ and so, by Sylow's Theorem, we can choose $g$ such that $\langle s,z \rangle^g =\langle t,z \rangle$.
Now $S^g$ and $S$ are Sylow $2$-subgroups of $N_G(\langle t,z \rangle)$, so by Sylow's Theorem there exists $x \in N_G(\langle t,z \rangle)$ with $S^{gx}=S$, and so $\langle s,z \rangle^h =\langle t,z \rangle$ with $h = gx \in N_G(S)$. Since $\langle s,z \rangle$ and $\langle t,z \rangle$ are the only two Klein $4$-subgroups of $S$, we must also have $\langle t,z \rangle^h =\langle s,z \rangle$.
So conjugation by $h$ is inducing an outer automorphism of $S$ of even order, which is impossible because $N_G(S)/S$ has odd order.
