Conjugacy Classes of the Quaternion Group $Q$ 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. 
 A: 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 $\{\pm k\}$.
A: Those elements commute with everything (they form the center) so they are singleton conjugacy classes.
A: it is quite a long time. However, after trying to find all conjugacy classes of Q8 for my homework, I share the following idea.
First, we know that $|C_{Q}(x)| = |Q|/|Cl(x)|$ where $Cl(x)$ is the conjugacy class of element $x$ and $C_Q(x)$ is the centralizer of $x$.
Let start with $i \in Q$, we have $\langle i \rangle \leq C_Q(i) \leq Q$. Note that $|\langle i \rangle| = 4$ and $C_Q(i) \neq Q$ then $|C_Q(i)| = 4$. It follows that $|Cl(i)| = 2$. We can also check that $-i \in Cl(i)$. Therefore,
$Cl(i) = \{i, -i\}$.
Similar argument works for conjugacy classes of $j$ and $k$, while $\{1\}$ and $\{-1\}$ are the remaining classes.
