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?
Best Answer
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.
In general however classifying groups and subgroups of finite order will require more machinery such as the Sylow Theorems and semidirect products. Counting arguments using these ideas are very powerful for classifying groups of arbitrary order and their subgroups.