To determine the commutator subgroup of the dihedral group $D_8$ and the quaternion group $Q_8$, let’s go through each case.

1. Commutator Subgroup of $D_8$

The dihedral group $D_8$ represents the group of symmetries of a square, including rotations and reflections, and has the following properties:

• Order: 8
• Elements: $D_8 = \{ e, r, r^2, r^3, s, rs, r^2s, r^3s \}$
  • Here, $r$ represents a rotation by $90^\circ$ and $s$ represents a reflection.

To find the commutator subgroup $D_8'$, we calculate the set of all commutators in $D_8$: $[x, y] = x^{-1} y^{-1} xy$

Commutator Calculations for $D_8$:

1. For rotations $r$, $r^2$, and $r^3$, the commutators are trivial (i.e., they commute with each other).
2. For reflections like $s$ and products like $rs$, we compute specific commutators:
   • $[r, s] = r^{-1} s^{-1} r s = r^3 s r s = r^2$
   • Similarly, other commutators yield elements like $r^2$.

After computing all commutators, we find that $D_8' = \langle r^2 \rangle = \{ e, r^2 \}$, which is the cyclic subgroup generated by $r^2$.

Thus, the commutator subgroup of $D_8$ is $\{ e, r^2 \}$.

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2. Commutator Subgroup of $Q_8$

The quaternion group $Q_8$ has the following elements:

• Order: 8
• Elements: $Q_8 = \{ \pm 1, \pm i, \pm j, \pm k \}$
  • Here, $i, j, k$ are the quaternion units, satisfying $i^2 = j^2 = k^2 = -1$ and $ij = k$, $jk = i$, $ki = j$.

For $Q_8$, we compute the commutators in a similar way:

• Commutator $[i, j] = i^{-1} j^{-1} i j = (-i)(-j)ij = k^2 = -1$
• Similarly, all non-identity commutators in $Q_8$ yield $-1$.

Thus, the commutator subgroup $Q_8'$ is the subgroup generated by $-1$: $Q_8' = \langle -1 \rangle = \{ 1, -1 \}$

Therefore, the commutator subgroup of $Q_8$ is $\{ 1, -1 \}$.

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Summary

• Commutator subgroup of $D_8$: $\{ e, r^2 \}$
• Commutator subgroup of $Q_8$: $\{ 1, -1 \}$

Would you like further details on commutator subgroup derivations or specific steps?

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1. What is the significance of the commutator subgroup in group theory?
2. How does the structure of $D_8$ compare to other dihedral groups?
3. Why is the commutator subgroup of $Q_8$ related to the element $-1$?
4. Can we generalize commutator subgroups for other quaternion groups?
5. What other properties distinguish $D_8$ from $Q_8$?

Tip: In group theory, the commutator subgroup is always a normal subgroup and often reveals information about a group’s structure, such as whether it is abelian.