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Handout Monday 9/29
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.
We said that \((ab)^{-1} = b^{-1}a^{-1}\text{.}\) In an abelian group, this is also the same as \(a^{-1} b^{-1}\text{.}\) We often use additive notation for abelian groups. So this would be stated as \(-(a+b) = (-a) + (-b)\text{.}\)
Careful about negatives. \(-a\) is NOT the same as \(-1\cdot a\text{.}\) Even if we are in a ring, there is no guarantee that the ring contains 1 or \(-1\text{.}\)
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In fact, in any ring, \((-a)(-b) = ab\text{.}\) What does this mean? Can we prove it?
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It would be easier to first prove that \(-(ab) = (-a)b\text{.}\) We have done this already, but let’s review it.
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We are saying the the additive inverse of \(ab\) is \((-a)b\text{.}\) To verify this, we must show that \(ab + (-a)b = 0\text{.}\) Use the distributive property to “factor out” the \(b\text{.}\)
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But then, \(0b = ?\text{...}\) is this zero? Yes. Let’s prove this too!
