Skip to main content

Introduction to Abstract Algebra I MATH 321 Fall 2025

Print preview

Paper size

Handout Wednesday 10/29

Last time we considered how you can sometimes define an operation out of the set of cosets. The group is called a factor group or quotient group. We will write this as \(G/H\) and say “\(G\) mod \(H\text{,}\)” where \(G\) is the larger group and \(H\) is the subgroup. When this worked, the operation was defined as \(aHbH = abH\text{.}\) But it didn’t always work. Today, we will explore the properties of subgroups \(H\) that make this work.
🔗
Those subgroups \(H\) that work will be called normal subgroups. But that’s not how we define normal subgroup. We define it as a property of the subgroup, and then notice that this is exactly the property that allows us to properly well-define the coset operation.
  • Our definition is: a subgroup \(N\) of a group \(G\) is normal provided \(aN = Na\) for all \(a \in G\text{.}\) That is, left cosets give the same partition as right cosets.
    🔗
    🔗
  • Do this with the example \(H = \{(1), (12)\}\) that we looked at last time. Verify that left cosets are not the same as right cosets.
    🔗
    🔗
  • What if \(G\) is abelian? Well, then left cosets will be exactly the same as right cosets (even element by element). For non-abelian groups, we can sometimes have left-cosets be right-cosets, as in \(H = \{(1), (123), (132)\}\text{.}\)
    🔗
    🔗
  • The previous example suggests that any time \(H\) has index 2 in \(G\) (i.e., there are exactly two cosets), we will have \(H\) normal in \(G\text{.}\) You are asked to prove this as a bonus on your homework.
    🔗
    🔗
  • Deciding whether a subgroup is normal is important, and we will want to prove that subgroups are normal. What can we do if \(G\) is not abelian and \([G : H] \ne 2\text{?}\) It would be nice to have an easier test than writing out all the left and right cosets.
    🔗
    🔗
  • Here is a clue: what does it mean to say \(gN = Ng\text{?}\) It does not mean that \(gn = ng\) for all \(g \in G\text{,}\) \(n \in N\text{.}\) Rather, we have \(gn = n'g\text{.}\) But this means that \(gng\inv = n'\text{.}\)
    🔗
    🔗
  • In other words, if you do \(gng\inv\) you always get some element of \(N\text{.}\) We call \(gng\inv\) a conjugate of \(n\text{.}\)
    🔗
    🔗
  • So to show that \(N\) is a normal subgroup of \(G\text{,}\) we just need to show that every conjugate of every element of \(N\) is still in \(n\text{.}\) In other words, that \(N\) is closed under conjugates.
    🔗
    🔗
  • Important: the conjugates of \(n\) are the result of conjugating by all elements of \(G\text{.}\) It doesn’t help at all to check conjugates by elements just from \(N\) (these will always be in \(N\text{,}\) right?).
    🔗
    🔗
🔗

Example 3.

Let’s prove that the center of a group is always a normal subgroup. Recall \(Z(G) = \{x \in G \st xg = gx \text{ for all } g \in G\}\text{.}\)
  • First, we already know that \(Z(G)\) is a subgroup (or should we do this again?).
    🔗
    🔗
  • To prove that \(Z(G)\) is normal, we will show that it is closed under conjugates. So let \(n \in Z(G)\) and \(g \in G\text{.}\) What can we say about \(gng\inv\text{?}\)
    🔗
    🔗
  • Well, since \(g \in G\) and \(n \in Z(G)\text{,}\) we have \(gn = ng\text{.}\) So \(gng\inv = ngg\inv = n\text{.}\) So indeed, \(gng\inv \in Z(G)\text{.}\)
    🔗
    🔗
🔗
Finally, let’s think about the operation on the set of cosets.
  • We need \(aHbH = abH\text{.}\) Well, if \(H\) is normal, then we get \(aHbH = abHH\text{,}\) and surely \(HH = H\text{.}\)
    🔗
    🔗
  • This is very sloppy: we can’t treat the set \(H\) like an element. Well, not officially, but in practice, this is usually okay.
    🔗
    🔗
  • What is really going on? We need every element of \(aHbH\) to be an element of \(abH\text{.}\) The elements of \(aH bH\) have the form \(ah_1bh_2\text{.}\) Now we can use associativity to think of this as \(a(h_1b)h_2\text{.}\)
    🔗
    🔗
  • Since \(H\) is normal, we know that \(h_1b = bh_3\) for some \(h_3 \in H\text{.}\) So now our element is \(abh_3h_1\text{.}\)
    🔗
    🔗
  • But \(h_3h_1 = h_4\) for some \(h_4 \in H\text{.}\) Thus we have an element in the coset \(abH\text{.}\)
    🔗
    🔗
  • So this is exactly what we need.
    🔗
    🔗
🔗
If there is time, consider how the factor group of rotations in \(D_4\) shows us the coarse structure of the group \(D_4\text{.}\) Or how \(\Z/2\Z\) shows us the structure of evens and odds.
🔗
🔗
🔗
PrevTopNext