TITLE: Subgroups of order 8 in the quasidihedral group of order 16
QUESTION [5 upvotes]: Why are there only $3$ subgroups of order $8$ in the quasidihedral group $QD_{16}$ of order $16$? (I am not interested in drawing the lattice of subgroups, but rather an argument convincing one that there can be only $3$ subgroups of order $8$).

REPLY [4 votes]: Recall that $QD_{16}$ is given by the presentation
$$
\big\langle r,s \;\bigr|\; r^8=s^2=1,sr=r^3s \big\rangle.
$$
Any subgroup of order $8$ must be normal (having index two), so each is the kernel of some  epimorphism $QD_{16}\to \mathbb{Z}/2\mathbb{Z}$.  But there are only three such epimorphisms, namely
$$
(r,s) \mapsto (1,0),\qquad (r,s)\mapsto (0,1),\qquad\text{and}\qquad (r,s)\mapsto (1,1).
$$
The same argument shows that any group with $k$ generators has at most $2^k-1$ subgroups of index two.