# 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?

## 2 Answers

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.

• slick. I should study more about the classification of low order finite groups... – SZN Apr 27 '17 at 1:53
• Definitely saving this problem as one of my favorites. – user425349 Apr 27 '17 at 2:08

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 $\{\pm 1,\pm k\}$.

• I see. What makes a subgroup "proper"? I feel like we haven't discussed that in my class just yet, but we will soon get there. for clarification purposes, Lagrange's Theorem, in a nutshell, says that the order of a subgroup is a divisor of the order of a group correct? – user425349 Apr 27 '17 at 1:47
• Correct on Lagrange's theorem. A proper subgroup is just a subgroup not equal to the original group. It is a convenient way of getting away from trivial examples (e.g. "does a group $G$ always have a subgroup? "sure, $G$ is a subgroup of $G$! "ok, duh, but any proper subgroups?" "well, the identity is always a subgroup :3" "OK, proper, nontrivial subgroups?"). – SZN Apr 27 '17 at 1:50
• I see. Some clarification: what does it mean when a subgroup has, for example, index $3$, instead of $2$? I reading this link that I have hyperlinked They mentioned something about cosets. How would that be relevant here, if it is? – user425349 Apr 27 '17 at 1:55
• Fortunately, classifying all groups of order $4$ is pretty easy :-) I'll add a proof sketch for completeness. – Bungo Apr 27 '17 at 2:12
• $7 \text{ choose } 3$ equals $35$, not $7$ – user133281 Apr 27 '17 at 7:26