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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?

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  • $\begingroup$ 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? $\endgroup$ – CyclotomicField Sep 29 '17 at 0:04
  • $\begingroup$ Not yet. Covering that stuff later. $\endgroup$ – SS' Sep 29 '17 at 0:12
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    $\begingroup$ 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. $\endgroup$ – Joppy Sep 29 '17 at 0:17
  • $\begingroup$ ok I'll answer with that in mind. $\endgroup$ – CyclotomicField Sep 29 '17 at 0:22
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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.

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    $\begingroup$ 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. $\endgroup$ – user441558 Sep 29 '17 at 1:17
  • $\begingroup$ @hellotinfish Yes, of course. I only included them as a general tool since they asked for a "methodical way to identify the subgroups". $\endgroup$ – CyclotomicField Sep 29 '17 at 19:30

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