Let {+1, i, j, k} be the quaternion group of order 8. Determine the structure of the inner automorphism group Inn(Q8). Show that S4 is isomorphic to Aut(Q8). Define the function @: Aut(Q8) -> S4 as follows: @: (1,2) -> P1, (1,3) -> P2, (1,4) -> P3, (2,3) -> P4, (2,4) -> P5, (3,4) -> P6, where each +: Q8 is the automorphism determined by: P1(i) = j, P2(i) = k, P3(i) = -k, P4(j) = i, P5(j) = -i, P6(j) = -j, the remaining elements of S4 in such a way that 48 isomorphisms can be defined. Which elements of S4 belong to 0-“(Inn(Q8))? Identify the factor group Aut(Q8)/ Inn(Q8).
The Correct Answer and Explanation is:
The problem revolves around the quaternion group Q8={1,i,j,k,−1,−i,−j,−k}Q_8 = \{1, i, j, k, -1, -i, -j, -k\}Q8={1,i,j,k,−1,−i,−j,−k}, a group of order 8. You are tasked with investigating the structure of its inner automorphism group Inn(Q8)\text{Inn}(Q_8)Inn(Q8), proving that Aut(Q8)\text{Aut}(Q_8)Aut(Q8) is isomorphic to S4S_4S4, and finding specific properties related to the factor group Aut(Q8)/Inn(Q8)\text{Aut}(Q_8)/\text{Inn}(Q_8)Aut(Q8)/Inn(Q8).
1. Inner Automorphism Group of Q8Q_8Q8:
The inner automorphism group Inn(Q8)\text{Inn}(Q_8)Inn(Q8) consists of automorphisms of the form:ϕg(x)=gxg−1for some g∈Q8 and x∈Q8.\phi_g(x) = gxg^{-1} \quad \text{for some } g \in Q_8 \text{ and } x \in Q_8.ϕg(x)=gxg−1for some g∈Q8 and x∈Q8.
We need to identify the elements of Inn(Q8)\text{Inn}(Q_8)Inn(Q8). The inner automorphisms of Q8Q_8Q8 correspond to conjugation by elements of Q8Q_8Q8. Since Q8Q_8Q8 has order 8, Inn(Q8)\text{Inn}(Q_8)Inn(Q8) is a subgroup of Aut(Q8)\text{Aut}(Q_8)Aut(Q8), and the order of Inn(Q8)\text{Inn}(Q_8)Inn(Q8) is the index of the center of Q8Q_8Q8.
- The center of Q8Q_8Q8 is Z(Q8)={±1}Z(Q_8) = \{ \pm 1 \}Z(Q8)={±1}, because 111 and −1-1−1 commute with all other elements of Q8Q_8Q8, while the other elements do not.
- The order of Z(Q8)Z(Q_8)Z(Q8) is 2, and the index of Z(Q8)Z(Q_8)Z(Q8) in Q8Q_8Q8 is 8/2=48/2 = 48/2=4.
Thus, the inner automorphism group Inn(Q8)\text{Inn}(Q_8)Inn(Q8) has order 4, which means it is isomorphic to Z2×Z2\mathbb{Z}_2 \times \mathbb{Z}_2Z2×Z2, a group of order 4.
2. Automorphism Group Aut(Q8)\text{Aut}(Q_8)Aut(Q8):
The group of automorphisms of Q8Q_8Q8, Aut(Q8)\text{Aut}(Q_8)Aut(Q8), consists of all bijections of Q8Q_8Q8 that preserve the group structure. It turns out that Aut(Q8)\text{Aut}(Q_8)Aut(Q8) is isomorphic to S4S_4S4, the symmetric group on 4 elements.
- The elements i,j,ki, j, ki,j,k in Q8Q_8Q8 generate the non-commutative part of the group, and automorphisms are determined by how they map. Since conjugation by elements in Q8Q_8Q8 induces specific permutations on i,j,ki, j, ki,j,k, the structure of Aut(Q8)\text{Aut}(Q_8)Aut(Q8) corresponds to the possible permutations of these three elements.
- The automorphisms of Q8Q_8Q8 can be seen to map {i,j,k}\{i, j, k\}{i,j,k} to each other in a way that preserves the group relations. There are 24 such automorphisms, which is the order of S4S_4S4, showing that Aut(Q8)≅S4\text{Aut}(Q_8) \cong S_4Aut(Q8)≅S4.
3. Function \@\@\@:
The function \@:Aut(Q8)→S4\@ : \text{Aut}(Q_8) \to S_4\@:Aut(Q8)→S4 defined by the relations in the question maps certain automorphisms to specific permutations. To explain the behavior of \@\@\@:
- \@\@\@ sends each automorphism to a permutation in S4S_4S4, specifically mapping the automorphisms determined by the images of iii and jjj to specific elements of S4S_4S4. For example, P1(i)=jP_1(i) = jP1(i)=j implies that the automorphism P1P_1P1 sends iii to jjj, and so on for other automorphisms.
- There are 48 distinct automorphisms of Q8Q_8Q8, so the map \@\@\@ can define a variety of isomorphisms from Aut(Q8)\text{Aut}(Q_8)Aut(Q8) to S4S_4S4.
4. Elements of S4S_4S4 in Inn(Q8)\text{Inn}(Q_8)Inn(Q8):
Since Inn(Q8)\text{Inn}(Q_8)Inn(Q8) is a subgroup of Aut(Q8)\text{Aut}(Q_8)Aut(Q8), it consists of automorphisms corresponding to conjugation by elements of Q8Q_8Q8. Therefore, the elements of S4S_4S4 that belong to Inn(Q8)\text{Inn}(Q_8)Inn(Q8) are those that correspond to the automorphisms defined by conjugation by the elements ±1\pm 1±1 and the non-central elements i,j,ki, j, ki,j,k.
5. Factor Group Aut(Q8)/Inn(Q8)\text{Aut}(Q_8)/\text{Inn}(Q_8)Aut(Q8)/Inn(Q8):
Finally, the factor group Aut(Q8)/Inn(Q8)\text{Aut}(Q_8)/\text{Inn}(Q_8)Aut(Q8)/Inn(Q8) is isomorphic to the quotient of S4S_4S4 by its subgroup isomorphic to Z2×Z2\mathbb{Z}_2 \times \mathbb{Z}_2Z2×Z2. This quotient group is isomorphic to S3S_3S3, since S4/(Z2×Z2)≅S3S_4 / (\mathbb{Z}_2 \times \mathbb{Z}_2) \cong S_3S4/(Z2×Z2)≅S3. This is the group of permutations of 3 elements, corresponding to the non-trivial automorphisms of Q8Q_8Q8 modulo the inner automorphisms.
Conclusion:
- Inn(Q8)≅Z2×Z2\text{Inn}(Q_8) \cong \mathbb{Z}_2 \times \mathbb{Z}_2Inn(Q8)≅Z2×Z2,
- Aut(Q8)≅S4\text{Aut}(Q_8) \cong S_4Aut(Q8)≅S4,
- Aut(Q8)/Inn(Q8)≅S3\text{Aut}(Q_8)/\text{Inn}(Q_8) \cong S_3Aut(Q8)/Inn(Q8)≅S3.
These results provide insight into the structure of the quaternion group and its automorphism group.
