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Introduction to Abstract Algebra I MATH 321 Fall 2025

Section Week 6 (9/29-10/3)

This week we continue investigating groups and rings and their properties in depth.
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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 \(-a\text{.}\)
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The multiplicative notation is nice because it allows us to use exponents, and get the usual rules for exponents. Mostly.
  • 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.
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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 \(-(a+b) = (-a) + (-b)\text{.}\)
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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 \(-1\text{.}\)
  • 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!
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  • We have \(0 + bb = bb = (0 + b)b = 0b + bb\text{.}\) But now add \(-(bb)\) to both sides.
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Handout Wednesday 10/1

Today in class we explored more proofs about groups and rings, including looking at the The Cancellation Property activity.
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Handout Friday 10/3

We explored more of the activity we started the previous time and concluded that in some rings, even if not every element has a multiplicative inverse (i.e., even if the ring is not a field), we can still cancel. In particular.
  • Some rings have the cancellation property: For all elements \(a, b, c\) in the ring with \(a \ne 0\text{,}\) if \(ab = ac\) then \(b=c\text{.}\)
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  • This is not true in rings like \(\Z_6\) though, since \(2\cdot 1 = 2 \cdot 4\text{,}\) for example.
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  • We also noticed that some rings have the zero product property: For all elements \(a, b\) in the ring, if \(ab = 0\) then either \(a=0\) or \(b=0\text{.}\)
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  • Again, \(\Z_6\) does not have this property, since \(2\cdot 3 = 0\) for example.
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  • Definitely all fields have both these properties. But also \(\Z\) and \(R[x]\) (as long as \(R\) has the properties).
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  • We proved that the two properties are in fact equivalent. This used the observation that \(ab = ac\) if and only if \(a(b-c) = 0\text{.}\)
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  • We call rings with these properties integral domains.
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  • Another way to describe integral domains are as commutative rings with unity that have no zero divisors, i.e., no element \(a \ne 0\) such that some \(b \ne 0\) exists with \(ab = 0\text{.}\) In other words, a zero divisor is a counterexample to the zero product property, so a ring with no zero divisors is a ring in which no elements disobey the zero-product property.
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