22. Let 0 G → H be a group homomorphism, and let K = ker = {a € G|0(a) = EH}. Show that gK = Kg for all g E G. (This exercise is referenced in Example 24.13.)
22. Let 0 G → H be a group homomorphism, and let K = ker = {a € G|0(a) = EH}. Show that gK = Kg for all g E G. (This exercise is referenced in Example 24.13.)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Number 22
![17. Consider again the subgroup H = {U, 4,
(a) Create the group table for Z12/H and verify that it's a group under coset addi-
tion.
(b) Find the order of each a + H E Z₁2/H. Is the group cyclic?
18. We've seen examples where the coset multiplication shortcut fails.
(a) With subgroup H = {e, u} of D4, we saw that rooH d'H # (r90 d')H. (See
Example 21.10.)
(b) With subgroup H = {e, h} of D4, we saw that rooH dH # (r90 d)H. (See
Exercise #11.)
In each of those cases, verify that the set inclusion (ab)HaH. bH still holds,
19. Let G be a group (not necessarily commutative), H a subgroup of G, and a, b E G.
Define the coset product by aH bH = {aß | a € aH, ß € bH}. Prove that
(ab)H C aH bH. (This exercise is referenced in the proof of Theorem 23.5. It is
also the statement of Theorem 24.2.)
20. Let G be a commutative group, H a subgroup of G, and a, b E G. Define the coset
product by aH bH = {a6|a E aH, ß e bH}. Prove that aH bH C (ab)H. (This
exercise is referenced in Example 24.3.)
Note: Combined with the result of Exercise #19, this shows that aH bH = (ab)H
when G is commutative, hence completing the proof of Theorem 21.8.
21. Consider the homomorphism 8 G(Z₁0)→ U₁0 where 8(a) = det a for all a E
G(Z10), and let K = ker & be the kernel of 8; i.e., K = {a E G(Z₁0) | 8(a) = 1}.
(a) Let a = [3] = G(Z10). Verify that a € K.
(b) Let g = [4] = G(Z₁0) so that ga is in the left coset gK. Find 6 € K for
which ga = ß.g.
(c) Explain why ga is also contained in the right coset Kg.
22. Let 0 G H be a group homomorphism, and let K = ker = {a = G|0(a)
= EH}. Show that gK = Kg for all g E G. (This exercise is referenced in Example
24.13.)
Hint: Be careful! 0(gk) = 0(kg) does not necessarily imply that gk = kg.
In
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Transcribed Image Text:17. Consider again the subgroup H = {U, 4,
(a) Create the group table for Z12/H and verify that it's a group under coset addi-
tion.
(b) Find the order of each a + H E Z₁2/H. Is the group cyclic?
18. We've seen examples where the coset multiplication shortcut fails.
(a) With subgroup H = {e, u} of D4, we saw that rooH d'H # (r90 d')H. (See
Example 21.10.)
(b) With subgroup H = {e, h} of D4, we saw that rooH dH # (r90 d)H. (See
Exercise #11.)
In each of those cases, verify that the set inclusion (ab)HaH. bH still holds,
19. Let G be a group (not necessarily commutative), H a subgroup of G, and a, b E G.
Define the coset product by aH bH = {aß | a € aH, ß € bH}. Prove that
(ab)H C aH bH. (This exercise is referenced in the proof of Theorem 23.5. It is
also the statement of Theorem 24.2.)
20. Let G be a commutative group, H a subgroup of G, and a, b E G. Define the coset
product by aH bH = {a6|a E aH, ß e bH}. Prove that aH bH C (ab)H. (This
exercise is referenced in Example 24.3.)
Note: Combined with the result of Exercise #19, this shows that aH bH = (ab)H
when G is commutative, hence completing the proof of Theorem 21.8.
21. Consider the homomorphism 8 G(Z₁0)→ U₁0 where 8(a) = det a for all a E
G(Z10), and let K = ker & be the kernel of 8; i.e., K = {a E G(Z₁0) | 8(a) = 1}.
(a) Let a = [3] = G(Z10). Verify that a € K.
(b) Let g = [4] = G(Z₁0) so that ga is in the left coset gK. Find 6 € K for
which ga = ß.g.
(c) Explain why ga is also contained in the right coset Kg.
22. Let 0 G H be a group homomorphism, and let K = ker = {a = G|0(a)
= EH}. Show that gK = Kg for all g E G. (This exercise is referenced in Example
24.13.)
Hint: Be careful! 0(gk) = 0(kg) does not necessarily imply that gk = kg.
In
t
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