Math 546
Homework 8
Due Wednesday, March 29.
Contents of this handout:
1. 546 Problems. All students (students enrolled in 546 and 701I) are required to attempt
these.
2. 701I Problems. Only students enrolled in 701I are required to attempt these. Students not
enrolled in 701I are welcome, of course, to attempt these as well.
3. Bonus Problems. All students (students in 546 and 701I) may turn these in for additional
credit.
4. Examples. I wrote solutions to certain non-assigned problems from the text. These may be
helpful for you when you are working on the problems.
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546 Problems.
9.5 Let G = Q8 = {±I, ±J, ±K, ±L} ⊂ GL(2, C) be the quaternions. Find the right cosets
of H in G for
a) H = hJi.
b) H = h−Ii.
9.6 Let G = D4 = {e, f, f 2 , f 3 , g, gf, gf 2 , gf 3 }, and let H = hgf 2 i. Recall, for all i, that
we have f i g = gf i .
• Find the right cosets of H in G.
• Find the left cosets of H in G.
9.8 Let G = Z4 × Z4 , and let H = h(0, 2)i = {(0, 0), (0, 2)}. Find the right cosets of H
in G.
9.12 Let G be a group. We define a relation on G for all a, b ∈ G by aRb if and only if there
exists x ∈ G such that a = xbx−1 . Prove that R is an equivalence relation on G.
Note: When aRb, we say that a and b are conjugates. The problem states that the relation of
conjugation on a group is an equivalence relation.
10.1 Let G = Q8 = {±I, ±J, ±K, ±L} ⊂ GL(2, C) be the quaternions. Find [G : H] for
each of
• H = h−Ii,
• H = hKi,
• H = h−Li.
10.2
a) Let G = Z48 , and let H = h32i ⊆ G. Find [G : H].
c) Let G = Z112 , and let H = h100i ⊆ G. Find [G : H].
10.3 b) Let G = Z6 × Z4 , and let H = h2i × h2i. Find [G : H].
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701I Problems.
9.15 Let G be a group, and define a relation on G × G by (a, b)R(c, d) if and only if ad = bc.
Give a condition on G which is equivalent to R being an equivalence relation on G × G.
Prove your answer.
Your solution should take the following form:
Claim: R is an equivalence relation on G × G if and only if
(The blank should contain the appropriate condition on G.)
.
Proof: (Here is where you prove the claim.)
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Bonus Problems.
9.18 Let G be a group, and suppose that H and K are subgroups of G. Suppose further that
there exists x, y ∈ G such that Hx = Ky. Prove that H = K.
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Examples.
9.7 Find the right cosets of the subgroup H = h(1, 0)i = {(0, 0), (1, 0), (2, 0)} in
G = Z3 × Z2 .
Solution: We have
H =H + (0, 0) = {(0, 0), (1, 0), (2, 0)},
H + (0, 1) = {(0, 1), (1, 1), (2, 1)}.
It follows that the set of right cosets of H in G has size two: {H + (0, 0), H + (0, 1)}.
We may also write {(0, 0), (0, 1)} for this set.
9.9 Find the right cosets of H = h(1, 1)i in G = Z2 × Z4 .
Solution: We have
H =H + (0, 0) = {(0, 0), (1, 1), (0, 2), (1, 3)},
H + (1, 0) = {(1, 0), (0, 1), (1, 2), (0, 3)}.
It follows that the set of right cosets of H in G has size two: {H + (0, 0), H + (1, 0)}.
We may also write {(0, 0), (1, 0)} for this set.
9.13 Suppose that G is a group and that A and B are subgroups of G. We define a relation
R on G by
xRy ⇐⇒ there exists a ∈ A and b ∈ B such that x = ayb.
Prove that R is an equivalence relation on G.
Proof: It suffices to verify that R satisfies the three conditions for an equivalence relation.
• reflexivity. To see that R is reflexive, let x ∈ G. Then we have x = exe. It follows
that xRx, and hence, that R is reflexive.
• symmetry. To see that R is symmetric, let x, y ∈ G, and suppose that xRy. Then there
exists a, b ∈ G such that x = ayb. We have x = ayb implies that a−1 xb−1 = y. It
follows that yRx, and hence, that R is symmetric.
• transitivity. To see that R is transitive, let x, y, z ∈ G, and suppose that xRy and yRz.
Then there exists a, b, c, d ∈ G such that x = ayb and y = czd. We then have
x = ayb = a(czd)b = (ac)z(db),
which implies that xRz. We conclude that R is symmetric.
Having verified the necessary conditions, we see that R is an equivalence relation.
10.2 b) Let G = Z54 , and let H = h24i ⊆ G. Find [G : H].
54
54
=
= 9. Therefore, we have |H| = o(24) = 9.
gcd(54, 24)
6
|G|
54
We conclude that [G : H] =
=
= 6.
|H|
9
Solution: We have o(24) =
10.3 a) Let G = Z6 × Z4 , and let H = h0i × Z4 . Find [G : H].
Solution: We have |G| = 24 and |H| = 4. It follows that [G : H] =
24
|G|
=
= 6.
|H|
4