(A) All the Elements of G Are Powers of 2, Hence G Is Generated by 2, and Is Cyclic

Q2.

(a) All the elements of G are powers of 2, hence G is generated by 2, and is cyclic.

(c) The map f from (Z8 , +8) to (G, ×255) defined by f(n) = 2n obviously preserves the group operation from if a + b < 8; if a + b ≥ 8, it follows from 28 = 256 ≡ 1 (mod 255).

(b) As the generator 2 is of order 8, the subgroups are generated by 21 = 2, 22 = 4, 24 = 16, and 28 = 256 = 1.

Hence the subgroups of G are:

G = (1, 2, 4, 8, 16, 32, 64, 128)

(1. 4, 16, 64)

(1, 16)

(1)

Careful analysis and inspection of the group table shows that the following map takes Z23 to G

(0, 0, 0) / e
(0, 0, 1) / a
(0, 1, 0) / b
(0, 1, 1) / g
(1, 0, 0) / c
(1, 0, 1) / h
(1, 1, 0) / d
(1, 1, 1) / f

(a) Follows from the group table; a2 = e, so a generates H = (e, a).

(b) Inspection of the representation as Z23 gives the cosets as:

(e, a)

(b, g)

(c, h)

(d, f)

(d) Inspection of the representations gives (e, b, c, d) as a subgroup of K of G such that:

G = H ⊗ K.

I’m not precisely sure what you’re asking here, but I think this gives K as an answer.

Q4 (not group theory)

Properties of complex numbers, among other things, leads to the following coordinate representation of the edges

Adding vectors, this leads to

(a) Inspection of this table leads to:

(b) Follows immediately from (a) and

we get

(d)

and

(a) By inspection, g= (135)(246); g2 = (153)(264); and h = (12)(45)(36).

(b) This depends on the order of multiplication; whether the group elements are considered to operate on the left or on the right of the vertices.

If right, then ghg−1 =(14)(23)(56), and it is the reflection about the line between the midpoints of sides 1 2 and 4 5, rotated anticlockwise by 2π/3, or the line between the midpoints of the sides 2 3 and 5 6.

If left, then ghg−1 =(16)(25)(34), and it is the reflection about the line between the midpoints of sides 1 2 and 4 5, rotated clockwise by 2π/3, or the line between the midpoints of the sides 1 6 and 3 4.

(c) They are not conjugate in G. G preserves pair of opposite vertices, and (16)(25)(34) exchanges only one pair of opposites, and (14)(25)(36) exchanges all three pairs of opposites.

(d) They are conjugate in S6. The map (46) clearly exchanges them.

(e). The conjugacy class of (26)(35) obviously (OK, it’s obvious if you embed G in the full group preserving the Euclidian plane) consists of all reflections about an axis containing a line connecting two opposite vertices, or,

(26)(35), (15)(24), and (13)(46).

(a) can be done by inspection of the products:

(b) The kernel of J is the collection of matrices of the form:

(d) Obviously, the image of J is such a matrix group: The collection of all the matrices of the form

Alternatively, again by inspection of the multiplication of the matrices in the image of J, the set of diagonal matrices also is isomorphic.