This is going to be the first time we step into some quasi-real mathematics. I'm going to give four properties of groups and their proofs. You might not think the proofs are relevant, but reading proofs and understanding them is generally a useful exercise. I say this because, in general, the beauty of mathematics comes from proofs and from the enormous amount of creative lengths to which the brain can stretch in order to solve a problem. The proofs I'm about to show you are not the result of creative genius. It is necessary, though, to read the simpler proofs in order to understand the harder ones. What can I say? Its a process.
Theorem: Uniqueness of Identity
In a group, G, there is one and only one identity element.
In a group, G, there is one and only one identity element.
Proof:
It is clear from the definition of a group that there exists at least one identity element of G. I now wish to show that the identity is unique. Suppose that both e and e' are identity elements of G. We know that ∀g∈G, g∙e = g and that ∀g'∈G, g'∙e' = g'. So then the choice of g = e' and g' = e gives us that e'∙e = e' and e∙e' = e. It then follows that both e and e' are equal to e'∙e, so e' = e and the identity is unique.
It is clear from the definition of a group that there exists at least one identity element of G. I now wish to show that the identity is unique. Suppose that both e and e' are identity elements of G. We know that ∀g∈G, g∙e = g and that ∀g'∈G, g'∙e' = g'. So then the choice of g = e' and g' = e gives us that e'∙e = e' and e∙e' = e. It then follows that both e and e' are equal to e'∙e, so e' = e and the identity is unique.
This proof demonstrates a very prolific proof technique. Almost any time you need to prove that something is unique, you choose two of whatever it is and show that they're equal. The property that identities are unique is neither surprising nor groundbreaking, but showing its truth is necessary and the world of group theory would be much different if it weren't true.
Theorem: Cancellation
If G is a group then ∀a,b,c∈G, b∙a = c∙a implies b = c and a∙b = a∙c implies b = c.
If G is a group then ∀a,b,c∈G, b∙a = c∙a implies b = c and a∙b = a∙c implies b = c.
Proof:
Suppose that b∙a = c∙a. Multiplication on the right by a-1 gives (b∙a)∙a-1 = (c∙a)∙a-1 and associativity then gives b∙(a∙a-1) = c∙(a∙a-1). From the definition of inverses, b = c as desired. A nearly identical argument gives us that a∙b = a∙c implies b = c (starting with multiplication on the left by a-1).
Suppose that b∙a = c∙a. Multiplication on the right by a-1 gives (b∙a)∙a-1 = (c∙a)∙a-1 and associativity then gives b∙(a∙a-1) = c∙(a∙a-1). From the definition of inverses, b = c as desired. A nearly identical argument gives us that a∙b = a∙c implies b = c (starting with multiplication on the left by a-1).
The property of cancellation is convenient, but is definitely not obvious. There exist many mathematical universes where it is not true that b∙a = c∙a implies b = c. The first such structure that I will show you (not any time soon, though) is a ring.
Theorem: Uniqueness of Inverses
If G is a group, then ∀a∈G there is a unique a-1∈G such that a∙a-1 = e = a-1∙a where e is the identity of G.
If G is a group, then ∀a∈G there is a unique a-1∈G such that a∙a-1 = e = a-1∙a where e is the identity of G.
Proof:
Choose an element a∈G. From the definition of a group there exists at least one inverse of a. To show that the inverse is unique, suppose that both b and c are inverses of G. Then a∙b = e and a∙c = e so a∙b = a∙c and canelation gives us that b = c so the inverse of a is unique, as desired.
Choose an element a∈G. From the definition of a group there exists at least one inverse of a. To show that the inverse is unique, suppose that both b and c are inverses of G. Then a∙b = e and a∙c = e so a∙b = a∙c and canelation gives us that b = c so the inverse of a is unique, as desired.
This is another useful property that may be intuitive, but requires proof. It does give us another common proof technique, though. You'll notice that all I did was start with an assumption and use all the definitions to get to my conclusion. Sometimes you don't need any tricks.
Theorem: Socks-Shoes Property
If G is a group then ∀a,b∈G, (a∙b)-1 = b-1∙a-1.
If G is a group then ∀a,b∈G, (a∙b)-1 = b-1∙a-1.
Proof:
Choose a,b∈G. From the definition of an inverse we note that (a∙b)∙(a∙b)-1 = e = (a∙b)-1∙(a∙b). However, we now note that (a∙b)∙(b-1∙a-1) = a∙(b∙b-1)∙a-1 = a∙a-1 = e, and (b-1∙a-1)∙(a∙b) = b-1∙(a-1∙a)∙b = b-1∙b = e. We then get that (a∙b)∙(b-1∙a-1) = e = (b-1∙a-1)∙(a∙b) and b-1∙a-1 is an inverse of a∙b but since inverses are unique, (a∙b)-1 = b-1∙a-1.
Choose a,b∈G. From the definition of an inverse we note that (a∙b)∙(a∙b)-1 = e = (a∙b)-1∙(a∙b). However, we now note that (a∙b)∙(b-1∙a-1) = a∙(b∙b-1)∙a-1 = a∙a-1 = e, and (b-1∙a-1)∙(a∙b) = b-1∙(a-1∙a)∙b = b-1∙b = e. We then get that (a∙b)∙(b-1∙a-1) = e = (b-1∙a-1)∙(a∙b) and b-1∙a-1 is an inverse of a∙b but since inverses are unique, (a∙b)-1 = b-1∙a-1.
We call this the Socks-Shoes Property because if you put your socks on and then you put your shoes on (like a∙b), then when you want to do the reverse process you have to take your shoes off before you take your socks off (like b-1∙a-1). This is a cool property, and is a little bit unexpected.
Next time we'll explore the wonderful world of subgroups.
References
Previous Related Post: Group Attributes
Text Reference: Gallain Chapter 2
Wikipedia: Group
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