Today we will continue to explore cosets and Lagrange’s Theorem. We should practice proving something is true about cosets. But then also, we will consider what a quotient group.
First, a quick example of Lagrange’s theorem. Suppose a subgroup of \(S_4\) contains all the 2-cycles and at least one 3-cycle. Must this be the entire group? Yes, because there are 6 2-cycles, and then also 4 elements that are the product of two disjoint 2-cycles. Any 3-cycle will get you it and its inverse. Plus the identity makes 13 elements. But the size of any subgroup must divide the size of the group, and \(S_4\) has size 24.
Another: Any group that has an element of order 3 and one of order 5 must have order a multiple of 15. This is because the element of order 3 generates a subgroup of size 3. The element of order 5 must generate a subgroup of size 5. The order of the group must be divisible by both 3 and 5, so must be divisible by 15.
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.
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 \(gng\inv = n'\text{.}\)
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?).
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, \(gng\inv \in Z(G)\text{.}\)
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 \(a(h_1b)h_2\text{.}\)
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.
As an example, we have the ring \(\Z\text{,}\) which is also an abelian group, and we formed the cosets \(\Z/5\Z\text{,}\) which we said was itself a group. But \(\Z\) is also a ring: multiplication makes sense. Can we also multiply cosets?
Back up a bit. What do cosets for rings even mean? We want to have a subring \(S\) (which is also a subgroup) of the ring \(R\text{.}\) Do we form a coset with \(a+S\) or \(aS\text{.}\) See why the latter doesn’t make sense.
Great, so we have cosets. Can we add and multiply them? Note that since the group part of a ring is always abelian, the subgroup will automatically be normal. So there will be no problem with coset addition. What about multiplication?
If we consider \((a+S)(b+S)\text{,}\) we want this to be \(ab+S\text{.}\) What are things inside \((a+S)(b+S)\text{?}\) It is some \((a+s_1)(b+s_2)\text{.}\) Simplify this.
An easy way to get an ideal in a ring is to look at \(\langle a \rangle\text{,}\) the set of all multiples of \(a\text{.}\) We call this the ideal generated by \(a\text{.}\)
