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Handout Friday 9/26
We now know what a group is and have some examples. Much of group theory is about deciding what properties must be true of all examples of groups. In other words, what do the group axioms guarantee must be true.
For example, all the groups we have considered have only one identity element. Every element in a group seems to have only one inverse. These are not axioms of the definition of a group. But must they be true of all groups anyway? We can prove they must be.
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First, prove that the identity element is unique. First maybe think about the Cayley table version.
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Suppose \(e\) and \(e'\) were both identity elements. Then \(ee' = e\) and \(e e' = e'\) so \(e = e'\text{.}\)
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If \(g\) is any element of a group \(G\text{,}\) then the inverse of \(g\) is unique. We denote the inverse \(g\inv\text{.}\)
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Let’s prove this. Say \(x_1\) and \(x_2\) are both inverses of \(g\text{.}\) What can we say about \((x_1 g)x_2\) and about \(x_1(g x_2)\text{?}\) By associativity, they must be the same. But one is \(x_2\) and the other is \(x_1\text{.}\)
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We also noticed that in any row of a Cayley table, no element appears more than once. What would this say in terms of equations?
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Well, look at row \(a\) and suppose some element \(g\) appears twice: in column \(b\) and in column \(c\text{.}\) This would mean that \(ab = ac\text{.}\) Can’t we just cancel the \(a\)’s?
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Yes we can! Multiply both sides by \(a\inv\) on the left.
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Let’s talk about inverses some more. One of the reading questions asked whether \((ab)\inv = a\inv b\inv\text{.}\) It doesn’t! But we do have the \((ab)\inv = b\inv a\inv\text{.}\) What does this even mean?
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This is not saying that you can distribute the exponents, as long as you are careful. This says that the inverse of the element \(ab\) is the element you get as the product of two inverses.
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It is helpful to establish the following lemma: if \(xy = e\) then \(x = y\inv\) and \(y = x\inv\text{.}\) In other words, the only way for the product of two elements to be the identity is if they are inverses of each other.
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To prove this, assume \(xy = e\text{.}\) Now multiply on the right by \(y\inv\text{.}\) Tada. Now start over and multiply on the left by \(x\inv\text{.}\)
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So why is \((ab)\inv = b\inv a\inv\text{?}\) Well, what is \((ab)(b\inv a\inv)\text{?}\) Use associativity.
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We also get that \((a\inv)\inv = a\) using the same trick.
A note about notation: we have been writing things as if the operation were multiplication. This is okay, but we need to remember that we are really using an arbitrary operation. We sometimes write the operation as addition, and then inverses are written as \(-a\text{.}\)
The multiplicative notation is nice because it allows us to use exponents, and get the usual rules for exponents. Mostly.
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Is \((ab)^2 = a^2b^2\text{?}\) Alas no. We would expect there to be problems because \((ab)^2 = abab\text{,}\) but \(a^2b^2 = aabb\text{.}\) We can’t switch the middle \(ba\) to \(ab\) unless the group is abelian.
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In fact, we can show that if \((ab)^2 = a^2b^2\) for all \(a, b \in G\text{,}\) then the group must be abelian. Do this.
