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Lehman MAT-314 Section 11: Normal Subgroups

- 1. Abstract Algebra, Saracino Second Edition Section 11 Normal Subgroups "In mathematics the art of proposing a question must be held of higher value than solving it." – Georg Cantor -
- 2. Lehman College, Department of Mathematics Galois and Abel Niels Henrik Abel (1802-1829) - Norwegian Mathematician Évariste Galois (1811-1832) - French Mathematician
- 3. Lehman College, Department of Mathematics Definition of Normal Subgroups (1 of 1) Definition: Let 𝐺 be a group. A subgroup 𝐻 of 𝐺 is said to be a normal subgroup of G if 𝑎𝐻 = 𝐻𝑎 for all 𝑎 ∈ 𝐺. Note: For any group 𝐺, 𝑒 and 𝐺 are normal subgroups since 𝑎 𝑒 = 𝑒 𝑎 = 𝑎 for any a ∈ 𝐺, and 𝑎𝐺 = 𝐺𝑎 = 𝐺. It follows directly from the definition that every subgroup of an abelian group is normal. However, the converse of the last statement is not true. We can show that every subgroup of the group of unit quaternions 𝒬8 is normal, but 𝒬8 itself is nonabelian. Note also that if 𝐻 is a normal subgroup of a group 𝐺, then it does not always mean that 𝑎ℎ = ℎ𝑎 for every ℎ ∈ 𝐻 and every 𝑎 ∈ 𝐺. This will be demonstrated in the following example.
- 4. Lehman College, Department of Mathematics Examples of Normal Subgroups (1 of 3) Example 1. Consider the group 𝐺 = 𝑆3: Let 𝐻 = 𝑒, 1 2 3 , (1 3 2) . Then 𝐻 is a subgroup of 𝐺. Since 𝑒, (1 2 3), and (1 3 2) are elements of 𝐻, then 𝑒𝐻 = 𝐻𝑒, 1 2 3 𝐻 = 𝐻(1 2 3), and 1 3 2 𝐻 = 𝐻(1 3 2). Now, Hence, 1 2 𝐻 = 𝐻(1 2). Again, Therefore 2 3 𝐻 = 𝐻 2 3 . 𝐺 = 𝑆3 = 𝑒, 1 2 , 1 3 , 2 3 , (1 2 3), (1 3 2) 1 2 𝐻 = 1 2 𝑒, 1 2 1 2 3 , 1 2 1 3 2 } = 1 2 , 2 3 , (1 3) 𝐻 1 2 = 𝑒 1 2 , 1 2 3 1 2 , 1 3 2 1 2 } = 1 2 , 1 3 , (2 3) 2 3 𝐻 = 2 3 𝑒, 2 3 1 2 3 , 2 3 1 3 2 } = 2 3 , 1 3 , (1 2) 𝐻 2 3 = 𝑒 2 3 , 1 2 3 2 3 , 1 3 2 2 3 } = 2 3 , 1 2 , (1 3)
- 5. Lehman College, Department of Mathematics Examples of Normal Subgroups (2 of 3) Example 1 (cont’d). Similarly, we can show that 1 3 𝐻 = 𝐻 1 3 . It follows that 𝐻 is a normal subgroup of 𝑆3. But for 1 2 3 ∈ 𝐺 and 2 3 ∈ 𝐺, we have: Hence, 2 3 1 2 3 ≠ 1 2 3 2 3 . Now, consider the subgroup 𝐾 = 𝑒, (1 2) of 𝐺 = 𝑆3. For this subgroup: Hence, 1 3 𝐻 ≠ 𝐻(1 3), and so 𝐾 = 𝑒, (1 2) is not a normal subgroup of 𝐺 = 𝑆3 2 3 1 2 3 = (1 3) 1 2 3 2 3 = (1 2) 1 3 𝐻 = 1 3 𝑒, (1 3)(1 2) = 1 3 , (1 2 3) 𝐻 1 3 = 𝑒 1 3 , (1 2) 1 3 = 1 3 , (1 3 2)
- 6. Lehman College, Department of Mathematics Examples of Normal Subgroups (3 of 3) Example 2. Recall the group 𝐺𝐿 𝑛, ℝ of 𝑛 × 𝑛 real matrices with nonzero determinant. The set S𝐿 𝑛, ℝ of matrices in 𝐺𝐿 𝑛, ℝ with determinant 1 is a subgroup of 𝐺𝐿 𝑛, ℝ . First, S𝐿 𝑛, ℝ is a nonempty subset of 𝐺𝐿 𝑛, ℝ , because 𝐼 ∈ S𝐿 𝑛, ℝ . Next, if A, B ∈ 𝑆𝐿 𝑛, ℝ then det 𝐴𝐵 = det 𝐴 det 𝐵 = 1 and det 𝐴−1 = det 𝐴 −1 = 1. So 𝑆𝐿 𝑛, ℝ is closed under both multiplication and inverses. In addition since if A ∈ 𝑆𝐿 𝑛, ℝ and 𝐵 ∈ 𝐺𝐿 𝑛, ℝ then Above we used the fact that det 𝐴𝐵 = det 𝐴 det 𝐵 and det 𝐴−1 = det 𝐴 −1 from linear algebra. S𝐿 𝑛, ℝ ⊴ 𝐺𝐿 𝑛, ℝ det 𝐵𝐴𝐵−1 = det 𝐵 det 𝐴 det 𝐵−1 = 1
- 7. Lehman College, Department of Mathematics Normal Subgroups (1 of 8) Theorem 1. Let 𝐺 be a group and let 𝐻 be a subgroup of 𝐺. The following conditions are equivalent: a) 𝐻 is a normal subgroup of 𝐺 b) 𝑔𝐻𝑔−1 ⊆ 𝐻 for all 𝑔 ∈ 𝐺 c) 𝑔𝐻𝑔−1 = 𝐻 for all 𝑔 ∈ 𝐺 Proof. 𝑎) ⇒ 𝑏) Let 𝑔 ∈ 𝐺. Since 𝐻 is a normal subgroup of 𝐺, then 𝑔𝐻 = 𝐻𝑔. Hence for any ℎ ∈ 𝐻, 𝑔ℎ ∈ 𝐻𝑔, so 𝑔ℎ = ℎ′ 𝑔 for some ℎ′ ∈ 𝐻. Consequently, we have 𝑔ℎ𝑔−1 = ℎ′ 𝑔𝑔−1 = ℎ′ ∈ 𝐻, and 𝑔𝐻𝑔−1 ⊆ 𝐻. 𝑏) ⇒ 𝑐) Let ℎ ∈ 𝐻, then 𝑔−1 ℎ 𝑔−1 −1 = 𝑔−1 ℎ𝑔 ∈ 𝐻.
- 8. Lehman College, Department of Mathematics Normal Subgroups (2 of 8) Theorem 1 (cont’d). 𝑏) ⇒ 𝑐) Let ℎ ∈ 𝐻, then 𝑔−1 ℎ 𝑔−1 −1 = 𝑔−1 ℎ𝑔 ∈ 𝐻. This means that we have 𝑔 𝑔−1 ℎ𝑔 𝑔−1 ∈ 𝑔𝐻𝑔−1 or ℎ ∈ 𝑔𝐻𝑔−1 . Hence, it follows that 𝐻 ⊆ 𝑔𝐻𝑔−1 , and 𝑔𝐻𝑔−1 = 𝐻. 𝑐) ⇒ 𝑎) Since 𝑔𝐻𝑔−1 = 𝐻 for all 𝑔 ∈ 𝐺, then 𝑔𝐻 = 𝐻𝑔 for all 𝑔 ∈ 𝐺. Hence, 𝐻 is a normal subgroup of 𝐺. ∎ Theorem 11.2. Let 𝐺 be a group, then any subgroup of the center 𝑍(𝐺) is a normal subgroup of 𝐺. Proof. Let 𝐻 be any subgroup of 𝑍(𝐺). Let ℎ ∈ 𝐻 and let 𝑔 ∈ 𝐺. Since we have 𝐻 ⊆ 𝑍(𝐺) then 𝑔ℎ = ℎ𝑔 and 𝑔ℎ𝑔−1 = ℎ𝑔𝑔−1 = ℎ ∈ 𝐻. It follows that 𝑔𝐻𝑔−1 ⊆ 𝐻 and 𝐻 is normal in 𝐺, by Theorem 1. ∎
- 9. Lehman College, Department of Mathematics Normal Subgroups (3 of 8) Example 2. Let 𝐺 be a group and let 𝐻 be a subgroup of 𝐺. If for all 𝑎, 𝑏 ∈ 𝐺, 𝑎𝑏 ∈ 𝐻 implies 𝑏𝑎 ∈ 𝐻, show that 𝐻 is normal in 𝐺. Solution. Let ℎ ∈ 𝐻 be arbitrary. Then for any 𝑔 ∈ 𝐺 Since 𝑎𝑏 ∈ 𝐻 implies 𝑏𝑎 ∈ 𝐻 for any 𝑎𝑏 ∈ 𝐻, It follows that 𝑔ℎ 𝑔−1 = 𝑔ℎ𝑔−1 ∈ 𝐻 for all ℎ ∈ 𝐻 and for all 𝑔 ∈ 𝐺. This means that 𝑔𝐻𝑔−1 ⊆ 𝐻 and 𝐻 is normal in 𝐺, by Theorem 1. ∎ ℎ = 𝑒ℎ = 𝑔−1 𝑔 ℎ = 𝑔−1 𝑔ℎ ∈ 𝐻
- 10. Lehman College, Department of Mathematics Normal Subgroups (4 of 8) Corollary 1. If 𝐺 is abelian then every subgroup of 𝐺 is normal. Proof. If 𝐺 is abelian then 𝑍 𝐺 = 𝐺 and every subgroup of 𝐺 is normal by Theorem 11.3. ∎ Theorem 11.3. Let 𝐺 be a group and let 𝐻 be a subgroup of 𝐺 such that 𝐺: 𝐻 = 2, then 𝐻 is normal in 𝐺. (index 2 subgroups are normal). Proof. Since there are exactly two distinct right cosets of 𝐻 in 𝐺, one of them is 𝐻 and the other must be 𝐺 − 𝐻 (set difference) since right cosets of 𝐻 partition 𝐺. Likewise, the distinct left cosets must be 𝐻 and 𝐺 − 𝐻.
- 11. Lehman College, Department of Mathematics Normal Subgroups (5 of 8) Proof (cont’d). Thus, the left cosets are the right cosets, and for any 𝑔 ∈ 𝐺, we have 𝑔𝐻 = 𝐻𝑔. ∎ Example 3. We know that every subgroup of an abelian group is normal. Is the converse true? That is, if every subgroup of a group is normal, is the group abelian? Solution. Consider the group of unit quaternions, where 𝒬8 = 1, −1, 𝑖, −𝑖, 𝑗, −𝑗, 𝑘, −𝑘 under multiplication, with 𝑖2 = 𝑗2 = 𝑘2 = 𝑖𝑗𝑘 = −1. The subgroups of 𝒬8 are: 1 , −1 , 𝑖 , 𝑗 , 𝑘 , and 𝒬8. It follows that 𝒬8 is a normal subgroup of itself. The subgroups 𝑖 , 𝑗 , and 𝑘 all have order 4, so they are index 2 subgroups and are therefore normal by Theorem 11.3.
- 12. Lehman College, Department of Mathematics Normal Subgroups (6 of 8) Solution (cont’d). The subgroups 1 and −1 are both contained in the center 𝑍 𝒬 and are therefore normal by Theorem 11.2. Thus, every subgroup of 𝒬8 is normal, but 𝒬8 itself is nonabelian. Example 4. We have 𝑆 𝑛 = 𝑛! and 𝐴 𝑛 = 𝑛!/2 so the index 𝑆 𝑛: 𝐴 𝑛 = 2 and 𝐴 𝑛 is a normal subgroup of 𝑆 𝑛for all 𝑛 by Theorem 11.3. Theorem 11.4. Let 𝐺 be a group and let 𝐻 be a subgroup of 𝐺. Suppose 𝑔 ∈ 𝐺, then 𝑔𝐻𝑔−1 is a subgroup of 𝐺 with the same number of elements as 𝐻. Proof. Let 𝑔 ∈ 𝐺, then 𝑔𝐻𝑔−1 ⊆ 𝐺. Now, 𝑔𝐻𝑔−1 ≠ ∅
- 13. Lehman College, Department of Mathematics Normal Subgroups (7 of 8) Proof (cont’d). Now, 𝑔𝐻𝑔−1 ≠ ∅ since we have 𝑔𝑒𝑔−1 = 𝑔𝑔−1 = 𝑒 ∈ 𝑔𝐻𝑔−1 . We will now show that 𝑔𝐻𝑔−1 is closed under multiplication. Let 𝑔ℎ1 𝑔−1 and 𝑔ℎ2 𝑔−1 be elements of 𝑔𝐻𝑔−1 for some ℎ1, ℎ2 ∈ 𝐻. Then since ℎ1ℎ2 ∈ 𝐻 because 𝐻 is a subgroup of 𝐺. For closure under inverses, let 𝑔ℎ𝑔−1 ∈ 𝑔𝐻𝑔−1 for some ℎ ∈ 𝐻. Then since ℎ−1 ∈ 𝐺 because 𝐻 is a subgroup of 𝐺. It follows that 𝑔𝐻𝑔−1 is a subgroup of 𝐺. 𝑔ℎ1 𝑔−1 𝑔ℎ1 𝑔−1 = 𝑔ℎ1 𝑔−1 𝑔 ℎ1 𝑔−1 = 𝑔ℎ1ℎ2 𝑔−1 ∈ 𝑔𝐻𝑔−1 𝑔ℎ𝑔−1 −1 = 𝑔−1 −1 ℎ−1 𝑔−1 = 𝑔ℎ−1 𝑔−1 ∈ 𝑔𝐻𝑔−1
- 14. Lehman College, Department of Mathematics Normal Subgroups (8 of 8) Proof (cont’d). To prove that |𝑔𝐻𝑔−1 | = |𝐻| we will construct a one-to-one and onto function from 𝐻 to 𝑔𝐻𝑔−1 . Let 𝑓: 𝐻 → 𝑔𝐻𝑔−1 be given by 𝑓 ℎ = 𝑔ℎ𝑔−1 for all ℎ ∈ 𝐻. To show that 𝑓 is one-to-one, let ℎ1, ℎ2 ∈ 𝐻 and let 𝑓 ℎ1 = 𝑓(ℎ2) then 𝑔ℎ1 𝑔−1 = 𝑔ℎ2 𝑔−1 . By left and right cancellation, ℎ1 = ℎ2 and 𝑓 is one-to-one. To show that 𝑓 is onto 𝑔𝐻𝑔−1 , let 𝑔ℎ𝑔−1 be an arbitrary element of 𝑔𝐻𝑔−1 , then 𝑔ℎ𝑔−1 = 𝑓(ℎ) for some ℎ ∈ 𝐻. It follows that 𝑓 is onto 𝑔𝐻𝑔−1 , and |𝑔𝐻𝑔−1 | = |𝐻| ∎ Corollary 11.5. If 𝐺 is a group and 𝐻 a subgroup of 𝐺 and no other subgroup has the same number of elements as 𝐻, then 𝐻 is normal in 𝐺.
- 15. Lehman College, Department of Mathematics Quotient (Factor) Groups (1 of 5) Proof . For any 𝑔 ∈ 𝐺, 𝑔𝐻𝑔−1 is a subgroup of 𝐺 with the same number of elements as 𝐻 . By the hypothesis, 𝑔𝐻𝑔−1 = 𝐻, so by Theorem 1, 𝐻 is normal in 𝐺. ∎ Notation: If 𝐻 is a normal subgroup of a group 𝐺, we write: Notation: If 𝐻 ⊴ 𝐺, then 𝐺/𝐻 is defined as the set of all left (= right) cosets of 𝐻 in 𝐺. That is 𝐺/𝐻 = 𝑎𝐻 | 𝑎 ∈ 𝐺 Theorem 11.6. Let 𝐺 be a group and let 𝐻 be a normal subgroup of 𝐺, then 𝐺/𝐻 is a group under the operation ∗ defined below: For all 𝑎𝐻, 𝑏𝐻 ∈ 𝐺/𝐻, 𝑎𝐻 ∗ 𝑏𝐻 = (𝑎𝑏)𝐻 𝐻 ⊴ 𝐺
- 16. Lehman College, Department of Mathematics Quotient (Factor) Groups (2 of 5) Proof. We want to show that the operation ∗ is a well- defined binary operation on the set 𝐺/𝐻. That is, if we have 𝑎𝐻 = 𝑎1 𝐻 and 𝑏𝐻 = 𝑏1 𝐻, then Now 𝑎𝐻 = 𝑎1 𝐻 means 𝑎1 𝑎−1 ∈ 𝐻 and 𝑏𝐻 = 𝑏1 𝐻 means 𝑏1 𝑏−1 ∈ 𝐻. It follows that we want: That is, we want 𝑎1 𝑏1 𝑎𝑏 −1 ∈ 𝐻. But 𝑎𝐻 ∗ 𝑏𝐻 = 𝑎1 𝐻 ∗ 𝑏1 𝐻 𝑎𝐻 ∗ 𝑏𝐻 = 𝑎𝑏 𝐻 = 𝑎1 𝑏1 𝐻 = 𝑎1 𝐻 ∗ 𝑏1 𝐻 𝑎1 𝑏1 𝑎𝑏 −1 = 𝑎1 𝑏1 𝑏−1 𝑎−1 = 𝑎1 𝑏1 𝑏−1 𝑎1 −1 𝑎1 𝑎−1 = 𝑎1 𝑏1 𝑏−1 𝑎1 −1 𝑎1 𝑎−1
- 17. Lehman College, Department of Mathematics Quotient (Factor) Groups (3 of 5) Proof (cont’d). We have Since 𝑎1 𝑎−1 ∈ 𝐻 and 𝑏1 𝑏−1 ∈ 𝐻, then 𝑎1 𝑏1 𝑎𝑏 −1 is an element of 𝐻, if and only if 𝑎1 𝐻𝑎1 −1 is a subset of 𝐻. That is, if 𝐻 is normal in 𝐺. We will now check if ∗ is associative. Let 𝑎𝐻, 𝑏𝐻, 𝑐𝐻 be elements of 𝐻/𝐺, then Hence, ∗ is an associative binary operation on 𝐺/𝐻. 𝑎1 𝑏1 𝑎𝑏 −1 = 𝑎1 𝑏1 𝑏−1 𝑎−1 = 𝑎1 𝑏1 𝑏−1 𝑎1 −1 𝑎1 𝑎−1 = 𝑎1 𝑏1 𝑏−1 𝑎1 −1 𝑎1 𝑎−1 𝑎𝐻 ∗ 𝑏𝐻 ∗ 𝑐𝐻 = 𝑎𝑏 𝐻 ∗ 𝑐𝐻 = 𝑎𝑏 𝑐𝐻 = 𝑎𝑏𝑐 𝐻 𝑎𝐻 ∗ 𝑏𝐻 ∗ 𝑐𝐻 = 𝑎𝐻 ∗ 𝑏𝑐 𝐻 = 𝑎 𝑏𝑐 𝐻 = 𝑎𝑏𝑐 𝐻
- 18. Lehman College, Department of Mathematics Quotient (Factor) Groups (4 of 5) Proof (cont’d). We will now check if ∗ has an identity element. Let 𝑎𝐻 be an element of 𝐻/𝐺, then So 𝑒𝐻 = 𝐻 is the identity element in 𝐺/𝐻 under ∗. For inverses, let 𝑎𝐻 ∈ 𝐺/𝐻, then It follows that 𝐺/𝐻 is closed under inverses and 𝐺/𝐻 is a group under the binary operation ∗. ∎ Definition: Let 𝐺 be a group and let 𝐻 be a normal subgroup of 𝐺. Then the group 𝐺/𝐻 of all left cosets of 𝐻 in 𝐺 under the binary operation 𝑎𝐻 ∗ 𝑏𝐻 = (𝑎𝑏)𝐻 is called the quotient group or factor group of 𝐺 by 𝐻. 𝑎𝐻 ∗ 𝑒𝐻 = 𝑎𝑒 𝐻 = 𝑎𝐻 = 𝑒𝑎 𝐻 = 𝑒𝐻 ∗ 𝑎𝐻 𝑎𝐻 ∗ 𝑎−1 𝐻 = 𝑎𝑎−1 𝐻 = 𝑒𝐻 = 𝑎−1 𝑎 𝐻 = 𝑎−1 𝐻 ∗ 𝑎𝐻
- 19. Lehman College, Department of Mathematics Quotient (Factor) Groups (5 of 5) Definition: Let 𝐺 be a group and let 𝐻 be a normal subgroup of 𝐺. Then the group 𝐺/𝐻 of all left cosets of 𝐻 in 𝐺 under the binary operation 𝑎𝐻 ∗ 𝑏𝐻 = (𝑎𝑏)𝐻 is called the quotient group or factor group of 𝐺 by 𝐻. We defined [𝐺: 𝐻] (the index of 𝐻 and 𝐺) as the number of distinct left (or right) cosets. Since 𝐺/𝐻 is the set of all left (or right) cosets of a normal subgroup 𝐺, then: If the group 𝐺 is finite, then by Lagrange’s Theorem, we have 𝐺: 𝐻 = |𝐺| |𝐻| . It follows that for finite groups 𝐺: |𝐺/𝐻| = [𝐺: 𝐻] |𝐺/𝐻| = |𝐺| |𝐻|
- 20. Lehman College, Department of Mathematics Examples of Quotient Groups (1 of 3) Example 5. Let 𝐺 = ℤ6 = 0, 1, 2, 3, 4, 5 under addition modulo 6. Then 𝐺 is cyclic, hence abelian. From Corollary 1, we know that every subgroup of 𝐺 is normal. Let 𝐻 = 0, 3 be a subgroup of 𝐺. Then 𝐻 ⊴ 𝐺, and the distinct left (right) cosets of 𝐻 in 𝐺 are: 𝐻 = 0 + 𝐻 = 0, 3 = 3 + 𝐻, 1 + 𝐻 = 1, 4 = 4 + 𝐻, and 2 + 𝐻 = 2, 5 = 5 + 𝐻. We see that |𝐺/𝐻| = 3, and |𝐺| = 6 with |𝐻| = 2, so 𝐺 /|𝐻| = 6/2 = 3. 1 + 𝐻 and 2 + 𝐻 are inverses of each other. 1 + 𝐻 and 2 + 𝐻 both have order 3. This is actually the group ℤ3. 𝐺/𝐻 = 𝐻, 1 + 𝐻, 2 + 𝐻 1 + 𝐻 + 2 + 𝐻 = 3 + 𝐻 = 𝐻 1 + 𝐻 + 1 + 𝐻 + (1 + 𝐻 = 3 + 𝐻 = 𝐻
- 21. Lehman College, Department of Mathematics Examples of Quotient Groups (2 of 3) Example 6. In Example 1, we looked at 𝐺 = 𝑆3 We showed that the subgroup 𝐻 = 𝑒, 1 2 3 , (1 3 2) is normal in 𝐺. Let us look at the set of distinct cosets of 𝐻 in 𝐺: Here: We see that |𝐺/𝐻| = 2, and |𝐺| = 6 with |𝐻| = 3, so 𝐺 /|𝐻| = 6/3 = 2. Now, So 1 2 𝐻 is its own inverse. This is the group ℤ2. 𝐺/𝐻 = 𝐻, (1 2)𝐻 𝐺 = 𝑆3 = 𝑒, 1 2 , 1 3 , 2 3 , (1 2 3), (1 3 2) 1 2 𝐻 = 1 2 , 2 3 , (1 3) = 2 3 𝐻 = 1 3 𝐻 𝐻 = 𝑒𝐻 = 𝑒, 1 2 3 , (1 3 2) = 1 2 3 𝐻 = 1 3 2 𝐻 1 2 𝐻 ⋅ 1 2 𝐻 = 1 2 1 2 𝐻 = 𝑒𝐻 = 𝐻
- 22. Lehman College, Department of Mathematics Examples of Quotient Groups (1 of 3) Example 7. Consider the group ℤ under addition and consider the subgroup 3ℤ = … , −6, −3, 0, 3, 6, … . Since ℤ is abelian, we know from Corollary 1 that every subgroup of ℤ is normal. The left cosets of 3ℤ in ℤ are: The distinct left cosets are: 3ℤ, 1 + 3ℤ and 2 + 3ℤ. Then 0 + 3ℤ = … , −9, −6, −3, 0, 3, 6, 9, … = 3ℤ 1 + 3ℤ = … , −8, −5, −2, 1, 4, 7, 10, … 2 + 3ℤ = … , −7, −4, −1, 2, 5, 8, 11, … 3 + 3ℤ = … , −6, −3, 0, 3, 6, 9, 12, … = 3ℤ 4 + 3ℤ = … , −5, −2, 1, 4, 7, 10, 13, … = 1 + 3ℤ −1 + 3ℤ = … , −10, −7, −4, −1, 2, 5, 8, … = 2 + 3ℤ −2 + 3ℤ = … , −11, −8, −5, −2, 1, 4, 7, … = 1 + 3ℤ ℤ/3ℤ = 3ℤ, 1 + 3ℤ, 2 + 3ℤ
- 23. Lehman College, Department of Mathematics Corollaries (1 of 10) Example 8. Let 𝐺 be a group and let 𝐻 be a normal subgroup of 𝐺. Show that if 𝐺 is abelian then the quotient group 𝐺/𝐻 is also abelian. Solution. Let 𝑎𝐻 and 𝑏𝐻 be arbitrary elements of 𝐺/𝐻 for some 𝑎, 𝑏 ∈ 𝐺. Then But 𝑎𝑏 = 𝑏𝑎 for all 𝑎, 𝑏 ∈ 𝐺 since 𝐺 is abelian. Hence, Since 𝑎𝐻 ∗ 𝑏𝐻 = 𝑏𝐻 ∗ 𝑎𝐻 for all 𝑎𝐻, 𝑏𝐻 in 𝐺/𝐻 it follows that 𝐺/𝐻 is abelian. ∎ 𝑎𝐻 ∗ 𝑏𝐻 = (𝑎𝑏)𝐻 𝑎𝐻 ∗ 𝑏𝐻 = 𝑎𝑏 𝐻 = 𝑏𝑎 𝐻 = 𝑏𝐻 ∗ 𝑎𝐻
- 24. Lehman College, Department of Mathematics Corollaries (2 of 10) Example 9. Let 𝐺 be a group and let 𝐻 and 𝐾 be subgroups of 𝐺 with 𝐻 normal in 𝐺. Show that 𝐻 ∩ 𝐾 is a normal subgroup of 𝐾. Solution. We know that 𝐻 ∩ 𝐾 is a subgroup of 𝐺. Since 𝐻 ∩ 𝐾 is contained in both 𝐻 and 𝐾 then 𝐻 ∩ 𝐾 is a subgroup of 𝐻 and 𝐾, respectively. In order show that 𝐻 ∩ 𝐾 ⊴ 𝐾, it is sufficient to show that for all 𝑘 ∈ 𝐾: Let 𝑎 ∈ 𝐻 ∩ 𝐾, then 𝑎 ∈ 𝐻 and 𝑎 ∈ 𝐾.Hence, 𝑘𝑎𝑘−1 ∈ 𝐾 for all 𝑘 ∈ 𝐾. We want to show that 𝑘𝑎𝑘−1 ∈ 𝐻 ∩ 𝐾. Since 𝐻 ⊴ 𝐺, then 𝑔𝑎𝑔−1 ∈ 𝐻 for all 𝑎 ∈ 𝐻 and all 𝑔 ∈ 𝐺. But 𝑘 ∈ 𝐾 ⊆ 𝐺 so 𝑘 ∈ 𝐺. It follows that 𝑘𝑎𝑘−1 ∈ 𝐻. 𝑘 𝐻 ∩ 𝐾 𝑘−1 ⊆ 𝐻 ∩ 𝐾
- 25. Lehman College, Department of Mathematics Corollaries (3 of 10) Solution (cont’d). Now, 𝑘𝑎𝑘−1 ∈ 𝐾 and 𝑘𝑎𝑘−1 ∈ 𝐻 for all 𝑘 ∈ 𝐾. It follows that 𝑘𝑎𝑘−1 ∈ 𝐻 ∩ 𝐾, and we have 𝑘 𝐻 ∩ 𝐾 𝑘−1 ⊆ 𝐻 ∩ 𝐾 for all 𝑘 ∈ 𝐾, but this implies that 𝐻 ∩ 𝐾 is normal in 𝐾 by Theorem 1. Example 10. Let 𝐺 be an abelian group of odd order 𝑛 and let 𝐺 = 𝑔1, 𝑔2, 𝑔3, … , 𝑔 𝑛 = 𝑒 . Show that the product Solution. Since 𝐺 is of odd order, by Lagrange’s Theorem 𝐺 can have no element of order 2 (the order of an element must divide the order of the group). Thus, each nonidentity element is distinct from its respective inverse. Since 𝐺 is abelian, then the above product is a product of (𝑛 − 1)/2 copies of the identity. ∎ 𝑔1 𝑔2 𝑔3 ⋯ 𝑔 𝑛 = 𝑒
- 26. Lehman College, Department of Mathematics Corollaries (4 of 10) Theorem 3. Let 𝐺 be a group. Show that if 𝐺/𝑍(𝐺) is cyclic, then 𝐺 is abelian. Proof: Let 𝑍 𝐺 = 𝑍. We know that 𝑍 ⊴ 𝐺, so 𝐺/𝑍 is a group under the operation 𝑎𝑍 ∗ 𝑏𝑍 = 𝑎𝑏 𝑍 for all elements 𝑎, 𝑏 ∈ 𝐺. Since 𝐺/𝑍 is cyclic, then 𝐺/𝑍 = ⟨𝑔𝑍⟩ for some 𝑔𝑍 ∈ 𝐺/𝑍. Let 𝑎, 𝑏 ∈ 𝐺, then 𝑎𝑍, 𝑏𝑍 ∈ 𝐺/𝑍. Since 𝐺/𝑍 is cyclic then 𝑎𝑍 = 𝑔𝑍 𝑛 = 𝑔 𝑛 𝑍 and, also 𝑏𝑍 = 𝑔𝑍 𝑚 = 𝑔 𝑚 𝑍 for some 𝑛, 𝑚 ∈ ℤ. It follows that we have 𝑎 ∈ 𝑔 𝑛 𝑍 and 𝑏 ∈ 𝑔 𝑚 𝑍. Therefore, 𝑎 = 𝑔 𝑛 𝑧1 and 𝑏 = 𝑔 𝑚 𝑧2 for some 𝑧1, 𝑧2 ∈ 𝑍. Now, 𝑎𝑏 = (𝑔 𝑛 𝑧1) 𝑔 𝑚 𝑧2 = 𝑔 𝑛 𝑔 𝑚 𝑧1 𝑧2 = 𝑔 𝑛+𝑚 𝑧1 𝑧2 = 𝑔 𝑚+𝑛 𝑧2 𝑧1 = 𝑔 𝑚 𝑔 𝑛 𝑧2 𝑧1 = (𝑔 𝑚 𝑧2)(𝑔 𝑛 𝑧1) = 𝑏𝑎
- 27. Lehman College, Department of Mathematics Corollaries (5 of 10) Example 11. Let 𝐺 be a group and let 𝐻 ⊴ 𝐺. Suppose 𝐻 = 2. Show that 𝐻 ⊆ 𝑍 𝐺 . Solution: Since 𝐻 = 2, then 𝐻 = 𝑒, 𝑎 for some 𝑎 ∈ 𝐺, (𝑎 ≠ 𝑒). Since 𝐻 ⊴ 𝐺, then 𝑔𝐻𝑔−1 = 𝐻 for all 𝑔 ∈ 𝐺, but So 𝑔𝑎𝑔−1 = 𝑎 for all 𝑔 ∈ 𝐺. Therefore 𝑔𝑎 = 𝑎𝑔 for all 𝑔 ∈ 𝐺. It follows that 𝑎 ∈ 𝑍 𝐺 . Since 𝑒 ∈ 𝑍(𝐺), then Theorem 4. Let 𝐺 be a finite group and let 𝐻 and 𝐾 be subgroups of 𝐺. 𝑔𝐻𝑔−1 = 𝑔𝑒𝑔−1 , 𝑔𝑎𝑔−1 = 𝑒, 𝑔𝑎𝑔−1 = 𝑒, 𝑎 = 𝐻 𝐻 = 𝑒, 𝑎 ⊆ 𝑍(𝐺)
- 28. Lehman College, Department of Mathematics Corollaries (6 of 10) Theorem 4. Let 𝐺 be a finite group and let 𝐻 and 𝐾 be subgroups of 𝐺. a) If |𝐻| and |𝐾| are relatively prime (gcd 𝐻 , |𝐾| = 1), then 𝐻 ∩ 𝐾 = 𝑒 . b) If 𝐻 and 𝐾 are two distinct subgroups, both of prime order 𝑝, then 𝐻 ∩ 𝐾 = 𝑒 . Proof (a). Now, 𝐻 ∩ 𝐾 is a subgroup of 𝐺 (cf. Homework 5, #1). Since 𝐻 ∩ 𝐾 is contained in both 𝐻 and 𝐾, then 𝐻 ∩ 𝐾 is subgroup of both 𝐻 and 𝐾. By Lagrange’s Theorem, |𝐻 ∩ 𝐾| is a divisor of both 𝐻| and |𝐾|, but gcd 𝐻 , |𝐾| = 1, so 𝐻 ∩ 𝐾 = 1 and 𝐻 ∩ 𝐾 = 𝑒 .
- 29. Lehman College, Department of Mathematics Corollaries (7 of 10) Proof (b). Now, 𝐻 ∩ 𝐾 is a subgroup of 𝐻, which has prime order. By Lagrange’s Theorem, 𝐻 ∩ 𝐾 = 1 or 𝐻 ∩ 𝐾 = 𝐻 . Since 𝐻 ∩ 𝐾 ⊆ 𝐻, then 𝐻 ∩ 𝐾 = 𝑒 or 𝐻 ∩ 𝐾 = 𝐻. Since 𝐻 ∩ 𝐾 ⊆ 𝐾, if 𝐻 ∩ 𝐾 = 𝐻, then 𝐻 ⊆ 𝐾. But 𝐻 = 𝐾 and 𝐻 ≠ 𝐾, so there is some 𝑎 ∈ 𝐻 with 𝑎 ∉ 𝐾. Therefore 𝐻 cannot be a subset of 𝐾. It follows then that 𝐻 ∩ 𝐾 = 𝑒 . ∎ Theorem 5. Let 𝐺 be a group of order 𝑝𝑞 for distinct primes 𝑝 and 𝑞 (𝑝 < 𝑞). Then all proper subgroups of 𝐺 are cyclic and 𝐺 has an element of order 𝑝. Proof. By Lagrange’s Theorem, the order of all subgroups of 𝐺 is a divisor of 𝐺 = 𝑝𝑞. The only proper
- 30. Lehman College, Department of Mathematics Corollaries (8 of 10) Proof (cont’d). The only proper divisors of 𝑝𝑞 are 1, 𝑝, and 𝑞. So, if 𝐻 is a proper subgroup of 𝐺, then we have 𝐻 = 1, 𝑝, or 𝑞. If 𝐻 = 1, then 𝐻 = 𝑒 = ⟨𝑒⟩ is cyclic. Otherwise, if 𝐻 = 𝑝, or 𝑞 (primes) then 𝐻 is cyclic. Let 𝑚 be the maximum order of all elements of 𝐺. Then, by Lagrange’s Theorem 𝑚 = 1, 𝑝, 𝑞, or 𝑝𝑞. If 𝑚 = 𝑝𝑞, then 𝐺 has an element 𝑎 of order 𝑝𝑞. Thus, 𝐺 = ⟨𝑎⟩ is cyclic. Since 𝑎 𝑝𝑞 = 𝑒, then 𝑎 𝑞 𝑝 = 𝑒 and 𝑎 𝑞 is an element of order 𝑝 in 𝐺. If 𝑚 = 𝑝, then 𝐺 has an element of order 𝑝. However, if 𝑚 = 𝑞 then there exists 𝑏 ∈ 𝐺 of order 𝑞. Consider the cyclic subgroup 𝐻 of 𝐺 generated by 𝑏. 𝐻 = 𝑒, 𝑏, 𝑏2 , 𝑏3 , … , 𝑏 𝑞−1 . Every nonidentity element of 𝐻 has order 𝑞.
- 31. Lehman College, Department of Mathematics Corollaries (9 of 10) Proof (cont’d). From Theorem 4(b), if 𝐾 is another subgroup of 𝐺 of order 𝑞, then 𝐻 ∩ 𝐾 = 𝑒 . It follows that 𝐻 has 𝑞 − 1 elements of order 𝑞 that are not in 𝐾, and vice versa. Since 𝐺 has one element 𝑒 of order 1, Then 𝐺 can have at most 𝑝𝑞 − 1 elements of order 𝑞. It follows that 𝐺 can have at most 𝑝 distinct subgroups of order 𝑞. Thus, the maximum number of elements of order 𝑞 in 𝐺 is 𝑝 𝑞 − 1 . Since if 𝐺 had 𝑝 + 1 distinct subgroups of order 𝑞, it would have 𝑝 + 1 𝑞 − 1 elements of order 𝑞. Now, 𝑝 + 1 𝑞 − 1 = 𝑝𝑞 + 𝑞 − 𝑝 − 1 = 𝑝𝑞 − 1 + (𝑞 − 𝑝). But 𝑞 > 𝑝, so 𝑞 − 𝑝 > 0 and the number of elements of order 𝑞 would exceed 𝑝𝑞 − 1.
- 32. Lehman College, Department of Mathematics Corollaries (10 of 10) Proof (cont’d). But the total number of nonidentity elements in 𝐺 is 𝑝𝑞 − 1. We know that 𝐺 has at most 𝑝 𝑞 − 1 = 𝑝𝑞 − 𝑝 elements of order 𝑞. So, the difference: Since the smallest prime is 2. It follows that 𝐺 has at least one element of order 𝑝. 𝑝𝑞 − 1 − 𝑝𝑞 − 𝑝 = 𝑝 − 1 ≥ 1