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Thus there are 10 subgroups of D8: the trivial subgroup, the six cyclic subgroups {e, s, s2,s3},{e, s2},{e, rx},{e, ry},{e, rx+y}, and {e, rx−y}, the two subgroups {e, s2,rx,ry} and {e, s2,rx+y,rx−y}, and D8. (4b) Show that D8 is not isomorphic to Q8.
The subgroup is (up to isomorphism) cyclic group:Z4 and the group is (up to isomorphism) dihedral group:D8 (see subgroup structure of dihedral group:D8). The subgroup is a normal subgroup and the quotient group is isomorphic to cyclic group:Z2.
Note that D8 has eight elements. The center of D8 is {R0, R180} (check this). Thus the number of elements in D4/Z(D4) is four, and hence it is isomorphic either to Z4 or to Z2 ×Z2.
Solution: (b) D8 is the symmetry group of a square and as such a Sylow 2-subgroup of S4.
, which is abelian. See center of dihedral group:D8. , which is of prime order, hence its Frattini subgroup is trivial.
Because each automorphism either fixes r or is β composed with an automorphism that fixes r, there are at most 8 automorphisms of D8. Now D8 has a non-trivial center, and in fact its center must have order 2 because G/Z(G) cannot be cyclic unless G is an abelian group. The center of D8 is the group {1,r2 }.
Cayley’s theorem says that D8 is isomorphic to a subgroup of S8, and gives a homomorphism φ : D8 → S8 where φ(g) is the permutation of the elements of D8 that is given by left- multiplication by g, according to our labeling.
But we now know that D8 has five distinct subgroups of order 2, which shows that D8 cannot be isomorphic to Q8.
Small dihedral groups
D1 and D2 are the only abelian dihedral groups.
The lattice of subgroups of D8 is given on [p69, Dummit & Foote]. All order 4 subgroups and 〈r2〉 are normal. Thus all quotient groups of D8 over order 4 normal subgroups are isomorphic to Z2 and D8/〈r2〉 = {1{1,r2},r{1,r2},s{1,r2}, rs{1,r2}} ≃ D4 ≃ V4.
Since the former is 2 and the latter is 3, we see that the order of π(τ) is 1, i.e. τ ∈ kerπ = D8. This means that D8 contains all transpositions and hence is equal to S4, a contradiction. Thus D8 is not normal in S4.
These groups are not isomorphic to eachother because D8 has an element of order 8 whereas D4 ⊕ Z2 does not.
It turns out that up to isomorphism, there are exactly 5 groups of order 8.
By Lagrange’s Theorem, all the proper subgroups of D10 are cyclic. So the subgroups are: D10, 〈a〉, 〈b〉, 〈ab〉, 〈a2b〉, 〈a3b〉, 〈a4b〉, {1}. In particular, there are 8 subgroups. [Diagram of subgroup lattice omitted.]
First, I’ll write down the elements of D6: D6 = {1,x,x2,x3,x4,x5,y,xy,x2y,x3y,x4y,x5y | x6 = 1,y2 = 1,yx = x5y}. This group has order 12, so the possible orders of subgroups are 1, 2, 3, 4, 6, 12.
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