Subgroups and The Quaternion Group I'm working on a rather difficult question. Perhaps somewhat could shed some light as to what I'm suppose to do:
Question: How many subgroups of index $2$ does the quaternion group have and what are they?
We haven't had much practice with using the quaternion group at all. Although, we have been focusing more on the Klein-$4$ group but that's not the point. From what I understand, upon doing some internet browsing, that the quaternion group has order $8$ since
$$Q = \{\pm 1, \pm i, \pm j, \pm k\}.$$
Now, if I  were to have a subgroup, call it $H$, such that it has index $2$ in, say, group $G$, it would mean that half of the elements in $G$ lie in $H$. So am I suppose to find all the subgroups of the quaternion group, then chose which ones have index $2$? How would one begin to answer this question?
 A: Look at the orders of the elements of $Q$:


*

*$1$ has order $1$

*$-1$ has order $2$

*the other elements all have order $4$


Up to isomorphism there are only two types of subgroups of order $4$, namely $\mathbb Z_4$ or $\mathbb Z_2 \times \mathbb Z_2$. The latter has three elements of order $2$, which is impossible in $Q$. So any subgroup of order $4$ must be cyclic. By inspection, the only possibilities are $\langle i \rangle$, $\langle j \rangle$, and $\langle k \rangle$.

Edit to sketch a proof of the statement about the possible types of group of order $4$:
Suppose $G$ is a group of order $4$. If $G$ contains an element of order $4$, then $G$ is cyclic, hence isomorphic to $\mathbb Z_4$.
The only other possibility is that $G$ does not contain an element of order $4$, in which case all of its non-identity elements must have order $2$ (Lagrange). Let $a$ and $b$ be two such elements.
Then $A = \langle a \rangle$ and $B = \langle b \rangle$ are subgroups of order $2$, hence index $2$, hence they are both normal.
Thus $AB$ is also a subgroup, and it contains $a$, $b$, and the identity, so its order is at least $3$. By Lagrange, its order must be $4$, hence $G = AB$. Since $A$ and $B$ are both normal subgroups and $A \cap B = 1$, this is a direct product.
Thus we have exposed $G$ as the direct product $A \times B$. As $A$ and $B$ have order $2$, they are both isomorphic to $\mathbb Z_2$, which shows that $G$ is isomorphic to $\mathbb Z_2 \times \mathbb Z_2$ as claimed.
A: Since the quaternions are a finite group of order $8$, Lagrange's theorem implies the only possible proper subgroups of $Q_8$ are of orders $2$ and $4$. Furthermore the only subgroups of index $2$ are of order $4$. Since $1$ must be an element of any subgroup then there are only $7$ choose $3$ (= $35$) more possibilities. Almost certainly $-1$ is also an element of at least one index-2 subgroup. I would only know of checking the rest by hand.   
Three are clearly $\{\pm 1,\pm i\}$, $\{\pm 1,\pm j\}$, and $\{\pm 1,\pm k\}$.
