Asambasam@ksu math 222 lecture notes 1 Groups
1.10 Elementary Properties of Groups
Theorem 1.10.1
If 𝐺 is a group, then the Identity element in 𝐺 is unique.
Proof: Let 𝐺 have two identity elements 𝑒 and 𝑓 under ∗.
Then 𝑒 ∗ 𝑓 = 𝑒 (Since 𝑓 is an identity element)
Similarly, 𝑒 ∗ 𝑓 = 𝑒 (Since 𝑒 is an identity element)
Hence e = 𝑒 ∗ 𝑓 = 𝑓
⇒ 𝑒 = 𝑓 as required.∎
Theorem 1.10.2
Let 𝐺,∗ be a group, then every element 𝑎 ∈ 𝐺 has exactly one inverse i.e unique inverse.
Proof: Suppose 𝑏 and 𝑐 are inverses of 𝑎 ∈ 𝐺 and 𝑒 ∈ 𝐺 be the identity element, then
𝑎 ∗ 𝑏 = 𝑒 and 𝑎 ∗ 𝑐 = 𝑒
Now, 𝑏 =𝑏∗𝑒 = 𝑏∗𝑎 ∗𝑐 = 𝑒∗𝑐 =𝑐
⇒𝑏=𝑐
Therefore, 𝑎 has exactly one inverse.
Theorem 1.10.3
Let 𝐺,∗ be a group, then ∀ 𝑎 ∈ 𝐺, 𝑎−1 −1 = 𝑎.
Proof: next slide.
Asambasam@ksu math 222 lecture notes 2 Groups
Proof: Let 𝑎 ∈ 𝐺 and 𝑒 ∈ 𝐺 be the identity element, then 𝑎−1 ∗ 𝑎 = 𝑒
Multiplying 𝑎−1 −1 on both sides we get
𝑎−1 −1 ∗ 𝑎−1 ∗ 𝑎 = 𝑎−1 −1 ∗ 𝑒
𝑎−1 −1 ∗ 𝑎−1 ∗ 𝑎 = 𝑎−1 −1 By associative law
Since 𝑎−1 −1 ∗ 𝑎−1 = 𝑒 we have
𝑒 ∗ 𝑎 = 𝑎−1 −1
⇒ 𝑎 = 𝑎−1 −1 ∎
Theorem 1.10.4 [Reversal law of inverse of product]
Let 𝐺,∗ be a group, then 𝑎, 𝑏 ∈ 𝐺, we have 𝑎 ∗ 𝑏 −1 = 𝑏−1 ∗ 𝑎−1 .
Proof: Assignment
Theorem 1.10.5 [cancellation law]
Let 𝑎, 𝑏, 𝑐 ∈ (𝐺,∗) such that either 𝑎 ∗ 𝑏 = 𝑎 ∗ 𝑐 or 𝑏 ∗ 𝑎 = 𝑐 ∗ 𝑎, then 𝑏 = 𝑐.
Proof: Let 𝑒 be an identity element in 𝐺. Taking the equation 𝑎 ∗ 𝑏 = 𝑎 ∗ 𝑐 ∀ 𝑎, 𝑏, 𝑐 ∈ 𝐺.
Now if 𝑎 ∈ 𝐺 ⇒ 𝑎−1 ∈ 𝐺
⇒ 𝑎−1 ∗ 𝑎 ∗ 𝑏 = 𝑎−1 ∗ 𝑎 ∗ 𝑐
𝑎−1 ∗ 𝑎 ∗ 𝑏 = 𝑎−1 ∗ 𝑎 ∗ 𝑐 by associative property
𝑒∗𝑏 =𝑒∗𝑐
⇒ 𝑏=𝑐
Similarly taking 𝑏 ∗ 𝑎 = 𝑐 ∗ 𝑎 we can post multiply by 𝑎−1 on both sides to get
𝑏 ∗ 𝑎 ∗ 𝑎−1 = 𝑐 ∗ 𝑎 ∗ 𝑎−1 ⇒ 𝑏 ∗ 𝑒 = 𝑐 ∗ 𝑒 ⇒𝑏=𝑐 ∎
Remark 1.10.6: Cancellation laws only apply to Abelian (commutative) groups.
Asambasam@ksu math 222 lecture notes 3 Groups
Theorem 1.10.7 ii. ∀ 𝑎, 𝑏 ∈ 𝐺, 𝑏𝑎−1 ∈ 𝐺 since 𝑎−1 ∈ 𝐺, 𝑏 ∈ 𝐺
A finite semi-group (𝐺,∗) satisfying cancellation Let 𝑦 = 𝑏𝑎−1
laws is a group. 𝑦𝑎 = 𝑏𝑎−1 𝑎
Theorem 1.10.8 = 𝑏(𝑎−1 × 𝑎)
Let (𝐺,×) be a group and 𝑎, 𝑏 ∈ 𝐺, then = 𝑏×𝑒
i. The equation 𝑎𝑥 = 𝑏 has a unique solution in =𝑏
𝐺. 𝑦𝑎 = 𝑏
ii. The equation 𝑦𝑎 = 𝑏 has a unique solution in
∴ 𝑦 = 𝑏𝑎−1 is the solution of 𝑦𝑎 = 𝑏
𝐺.
Assume 𝑦1 and 𝑦2 are solutions to 𝑦𝑎 = 𝑏, then
Proof:
𝑦1 𝑎 = 𝑏 = 𝑦2 𝑎
i. Let 𝑎, 𝑏 ∈ 𝐺, then
⇒ 𝑦1 = 𝑦2 by right cancellation law.
𝑎−1 𝑏 ∈ 𝐺 ∀ 𝑎−1 ∈ 𝐺, 𝑏 ∈ 𝐺
Hence, the solution is unique.∎
Now let 𝑥 = 𝑎−1 𝑏
Theorem 1.10.9
⇒ 𝑎𝑥 = 𝑎 𝑎−1 𝑏
Let 𝐺,∗ be a group, then the unique solution of the
𝑎𝑥 = 𝑎 × 𝑎−1 𝑏 = 𝑒 × 𝑏 = 𝑏 group equation 𝑥 ∗ 𝑥 = 𝑥 is 𝑥 = 𝑒 ∀𝑥, 𝑒 ∈ 𝐺.
𝑎𝑥 = 𝑏 ⇒ 𝑥 = 𝑎−1 𝑏 is the solution of Proof
the equation 𝑎𝑥 = 𝑏.
Let 𝑥 ∈ 𝐺, ⇒ 𝑥 −1 ∈ 𝐺.
Let us assume that 𝑥1 and 𝑥2 be two solutions to the
equation 𝑎𝑥 = 𝑏. Now 𝑥 −1 ∗ (𝑥 ∗ 𝑥) = 𝑥 −1 ∗ 𝑥
⇒ 𝑎𝑥1 = 𝑏 and 𝑎𝑥2 (𝑥 −1 ∗ 𝑥) ∗ 𝑥 = 𝑒
⇒ 𝑎𝑥1 = 𝑎𝑥2 ⇒ 𝑥1 = 𝑥2 by left cancellation. 𝑒∗𝑥 =𝑒
Hence the solution is unique.∎ 𝑥 = 𝑒∎
