# Finding all the subgroups of the quaternions

Consider the quaternions group $Q$ consisting of the eight elements $\pm1, \pm i, \pm j \pm k$ such that $i^2 = j^2 = k^2 = -1$ and $ijk = -1$. Identify all the subgroups.

I was trying to identify all the subgroups of Q and show that the proper ones are cyclic, which I got to be:

$\{1\}, \{-1,1\}, \{-1,1,-i,i\}, \{-1,1,-j,j\}, \{-1,1,-k,k\}$, and $\{-1,1,-i,i,-j,j,-k,k\}$

Now I want to show each proper subgroup is cyclic. So for the first 5 subgroups listed, they are generated by $\langle 1\rangle , \langle -1\rangle, \langle i\rangle, \langle j\rangle, \langle k\rangle$ and as such are cyclic.

Is this correct? And is there a methodical way for identifying the subgroups?

• This looks correct. Are you familiar with Lagrange's theorem concerning the possible orders of such subgroups? Do you know anything about normal subgroups or conjugacy classes yet? – CyclotomicField Sep 29 '17 at 0:04
• Not yet. Covering that stuff later. – SS' Sep 29 '17 at 0:12
• Looks correct! You can apply a little reasoning to show you have them all: A nontrivial subgroup has a non-identity element, and must contain the cyclic subgroup generated by that element. Considering each of $-1$, $\pm i$, $\pm j$, $\pm k$ gets you those first four subgroups. Now you've exhausted subgroups generated by one element, move on to 2 elements. $1$ and $-1$ won't give you anything new ($1$ for obvious reasons, $-1$ since you've just shown that it lives in every subgroup), so consider a group generated by two elements like $i, j$: this should be the whole of $Q$. Then you're done. – Joppy Sep 29 '17 at 0:17
• ok I'll answer with that in mind. – CyclotomicField Sep 29 '17 at 0:22

Looks correct. By Lagrange's Theorem, a consequence of Cauchy's Theorem the order of a subgroup must divide the order of the group. Since the order of the quaternions is 8 this means any proper, nontrivial subgroup must be of order 4 or 2. You'll also find that transforming a subgroup $H$ by $gHg^{-1}$, a group action called conjugation by $g$ will preserve the subgroup structure, although it may contain different elements entirely. In this particular case this is enough information to classify all the subgroups. I'd encourage you to calculate $gHg^{-1}$ for the groups of order four with $g$ being in the set $\{i,j,k\}$ to gain intuition about inner automorphisms which will be important later for classifying groups.
• Unfortunately, Sylow theorems cannot be applied to $p$-groups such as the $Q_8$ concerned and a generalized quaternion group cannot be expressed as a semidirect product of two of its nontrivial proper subgroups. – user441558 Sep 29 '17 at 1:17