4
$\begingroup$

I am trying to study the quaternion group $Q =\{\pm1,\pm i,\pm j,\pm k\}$, where $i^2 = j^2 = k^2 = -1$, $ij = k$, $jk = i$, $ki = j$.

First, I'm trying to find the conjugacy classes of $Q$.

The conjugacy class defined for an element $a$ in $Q$ is $$ (a)=\{b = gag^{-1}\mid g\in Q\}. $$

I am trying $a=-i$ and found $$(-i)=\left\{i= \left\{ \begin{array}{c} j\\ -j\\ k\\ -k\\ 1\\ -1\\ \end{array}\right\} \cdot-i\cdot \left\{ \begin{array}{c} -j\\ j\\ -k\\ k\\ 1\\ -1\\ \end{array}\right\}\right\}$$

So, shouldn't the the elements $1$ and $-1$ follow also the rule to say that $(-i)=i$?

I am quite confused, any hint is appreciated.

$\endgroup$
4
$\begingroup$

Disclaimer: This is not exactly an explanation, but a relevant attempt at understanding conjugates and conjugate classes.


Here is the "conjugation table" of $Q$, where each element in the table is equal to the row header conjugated by the column header, and the end of each row is the conjugacy class of the row header.
The table shows that $1$ and $-1$ form separate conjugacy classes.

\begin{array}{|r|rr|rr|rr|rr|c|} \hline & 1 & -1 & i & -i & j & -j & k & -k & \text{Conjugacy Class} \\ \hline 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & \{1\} \\ \hline -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & \{-1\} \\ \hline i & i & i & i & i & -i & -i & -i & -i & \{i,-i\} \\ -i & -i & -i & -i & -i & i & i & i & i & \{i,-i\} \\ \hline j & j & j & -j & -j & j & j & -j & -j & \{j,-j\} \\ -j & -j & -j & j & j & -j & -j & j & j & \{j,-j\} \\ \hline k & k & k & -k & -k & -k & -k & k & k & \{k,-k\} \\ -k & -k & -k & k & k & k & k & -k & -k & \{k,-k\} \\ \hline \end{array}

Example: $j$ conjugated by $-i$ is $(-i)\cdot j\cdot(-i)^{-1} = -j$.

Might also be noteworthy, to speed up the calculation of conjugations inside $Q$:

  • First two rows. $1$ and $-1$ form their own conjugacy class because they commute with all elements in $G$: $$\forall g \,[g\in G \rightarrow [g\cdot g^{-1} = 1 \land g\cdot 1\cdot g^{-1} = 1 \land g \cdot (-1)\cdot g^{-1} = -1]].$$
  • First two columns. For the same reason, every element conjugated by $1$ or $-1$ is itself: $$\forall g \,[g\in G \rightarrow 1\cdot g\cdot 1^{-1} = (-1)\cdot g\cdot (-1)^{-1} = g].$$
  • Every two rows grouped. Negate a single factor, and you negate the entire product. Therefore, every such two rows are negation of one another: $$\forall g\forall h \,[(g\in G \land h\in G) \rightarrow -(g\cdot h\cdot g^{-1}) = g\cdot (-h)\cdot g^{-1}].$$
  • Every two columns grouped. Two negations make a positive in products, therefore every such two columns are equal, e.g. $$\forall g\forall h \,[(g\in G \land h\in G) \rightarrow g\cdot h\cdot g^{-1} = (-g)\cdot h\cdot (-g)^{-1}].$$

Therefore, one only needs to figure out the first two rows, first two columns, and each of the 9 cases where $i,j,k$ gets conjugated by $i,j,k$, to populate the entire conjugation table and obtain conjugacy classes.

In conclusion, there clearly are five conjugacy classes: $\{1\}$, $\{-1\}$, $\{\pm i\}$, $\{\pm j\}$ and $\{\pm k\}$.

$\endgroup$
  • $\begingroup$ The "conjugation table" is, in fact, the left action of $Q$ on $Q$ through conjugation: $Q\times Q\to Q$ and $(g,h)\mapsto ghg^{-1}$. The center of $Q$ is $\{\pm 1\}$. The orbit of $1$ in $Q$ is $\mathcal O_1=\{1\}$ and the orbit of $-1$ in $Q$ is $\mathcal O_{-1}=\{-1\}$. Both these orbits are trivial orbits. The $\{\pm i\}$, $\{\pm j\}$ and $\{\pm k\}$ are nontrivial orbits, which are called conjugacy classes because the group action is conjugation. $\endgroup$ – Frenzy Li Apr 7 '17 at 15:36
1
$\begingroup$

Those elements commute with everything (they form the center) so they are singleton conjugacy classes.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.