I need some solution verification on basic group theory. Let group $G$ be the quasidihedral group of order $16,$ i.e. the group presentation is:
$$G=\langle \sigma, \tau \mid \sigma^8=1, \tau^2=1, \sigma\tau=\tau\sigma^3 \rangle $$
Now let $H=\langle\sigma^4\rangle $. This group is a normal subgroup of $G$, so we can think of the quotient group $G/H$. I need to calculate the order of the element in the quotient group, $\bar{\tau}\bar{\sigma} \in G/H$. I calculated the order of this element as $$(\tau H\sigma H)^2=(\tau\sigma)^2H=\tau\sigma\tau\sigma H=\tau(\tau\sigma^3)\sigma H=\sigma^4 H$$ and as $\sigma^4 \in H$, the order of this element is 2.
Am I right? Are there any parts to be improved? Thanks in advance!