Course 111: Algebra, 22nd November 2006
1. Consider the set of elements
G = {0, 1, 2, 3, 4, 5}
which forms a cyclic group under addition modulo 6.
List the proper subgroups of G and the generators of the group.
(Recall that {G} and {e}, where e = 0 here, are improper subgroups)
Determine the left and right cosets of H = {0, 3} in G and construct
the Cayley table of the quotient group G|H.
Proper subgroups
{0, 3} and {0, 2, 4}.
Generators
1, 5 since all the group elements can be generated by repeated addition
of 1 to itself (mod 6) or 5 to itself (mod 6).
Left & right cosets of H = {0, 3} in G
{0, 3}, {1, 4}, {2, 5} and the left and right cosets are the same.
Cayley table
The group G|H is the set of left (or right) cosets = {{0, 3}, {1, 4}, {2, 5}}
and writing {0, 3} = H, {1, 4} = A, {2, 5} = B the Cayley table is
H
A
B
H
H
A
B
A B
A B
B H
H A
2. Recall the properties of cosets listed in your notes. From this list prove
the following:
For a subgroup H of a group G,
• x ∈ xH for all x ∈ G.
Proof: Let x ∈ G. Then x = xe for e the identity element of G.
But also, e ∈ H. Then it follows that x ∈ xH.
• if x and y ∈ G and if y = xa for some a ∈ H then xH = yH.
Proof: Let x, y ∈ G with y = xa for some element a ∈ H.
Then yh = x(ah) and xh = y(a−1 h) ∀h ∈ H. Also, ah ∈ H
and a−1 h ∈ H ∀h ∈ H because H a subgroup of G. Therefore,
yH ⊂ xH and xH ⊂ yH and so yH = xH.
• each left coset of H in G has the same number of elements as H.
Proof: Consider xH ∩ yH to be nonempty for some x, y ∈ G.
And z ∈ (xH ∩ yH). Then z = xa for some a ∈ H and z = yb
for some b ∈ H. Then from the second proof here, zH = xH and
zH = yH so xH = yH.