How many elements of order 2 are there in the group of order 16 generated by $a$ and $b$ such that the order of $a$ is 8 and order $b$ is 2 and $bab^{-1} = a^{-1}$
My approach:
Let G be the give group of order 16.
Let $H = <a>$, subgroup of G. In H, $a^4$ is order 2.
$b<a>$ are the remaining 8 elements.
Now the index of $H$ is 2.
Let $x\in b<a>$. clearly $b\not\in H$. $H$ and $Hx$ are two distinct cosets.
Then $Hx^2 = He = H$
So every element in $b<a>$ is of order $2$.
So total number of elements of order 2 is 9.
Ask:
Is my answer 9 correct. I don't know where to use the fact given in question that $bab^{-1} = a^{-1}$
If my answer is wrong please provide any hints or answer.
I am doing a graduation course.
Work with $G' = \Bbb Z_{16}$ and you will see how your argument fails instantly. $H' = \langle 2 \rangle$ is a subgroup of order $8$, and $H'$ and $H'x$ are the only cosets, and $H'x^2 = H$, but not every element in $H'x$ have order $2$.
Every element can be represented as either $a^k$ or $a^k b$. (why?)
The only element in $a^k$ of order $2$ is $a^4$ as you have shown, since $a$ has order $8$.
From $bab^{-1} = a^{-1}$ we obtain $ba = a^{-1}b$, whence $(a^k b)^2 = a^k b a^k b = a^k a^{-k} b b = e$, so every element of the form $a^k b$ is of order $2$.
Therefore, there are $9$ elements of order $2$.
[Note: this is the dihedral group $D_8$ or $D_{16}$ of order $16$.]