Finding the number of elements of order $2$ in a given group 
How many elements of order $2$ are there in the group of order $16$ generated by $a$ and $b$ such that $o(a)=8$ and $o(b)=2$ and $bab^{-1}=a^{-1}$?

The basic thing i do not understand is that order of the generator must be equal to the order of the group. So how can $a$ and $b$ in this case generate the given group ?
I tried something like this, as $b$ is of order $2$ it must be its own inverse. So from the given condition on $a$ and $b$ replacing $b$ by $b^{-1}$ i get that $o(ba)=2$ . SO there are at least two elements of order two. However, this answer is obviously not good enough.

It's a group, so there's some identity element $1$. The order of $a$ is $8$, so $a,a2,…,a8=1$ are all distinct. The order of $a^2$ is $4$ since $1=a^8=(a^2)^4$ but $a^4$ and $a^6$ are not $1$. Since the order of $b$ is $2$, $b≠a$ and $b≠a^2$. Continue this process and work out a multiplication table.
