In this lecture, we covered matrix permutations, simple groups, and how to enumerate the normal subgroups of a group. If you can improve these notes, please do.
1Office hours¶
Professor Darmon will be holding office hours next Monday from 1012.
2Matrix permutations¶
Last time, we saw that $S_4/V \cong S_3$ where $V$ is the group of 22 cycles (e.g. $(12)(34)$) plus, of course, the identity. We also started looking at an example with matrices: $GL_n(\mathbb R)$, the group of $n\times n$ matrices with real entries. The permutation $\delta = (12)$^{1} would, in this case, indicate transposing columns 1 and 2 (the first two) while keeping the rest constant. We could also express this as a function  for instance, $f: S_3 \to GL_3(\mathbb R)$ would be a permutation of the columns.
If we then look at the determinant of the resulting permuted matrices, we get a quantitative way of seeing the effects of permutations on orientation. Reflections, for example, are orientationreversing, while rotations preserve orientation. The function for computing the determinant is in fact a homomorphism, $g: GL_n(\mathbb R) \to \mathbb R^{\times}$ (can't have a determinant of 0 in this case) and so the composition $h=g \circ f$ would be a function $h: S_n \to \{1, +1\}$, which we can call the sign homomorphism (incidentally, a surjective homomorphism).
Consider, now, the kernel of $h$, which we will call $A_n$ for now. The cardinality of $S_n$ is $n!$, and the cardinality of the image of $h$ ($\{1, +1\}) is 2, so the cardinality of $A_n$ is $n!/2$. So $A_n$ (called the alternating group) consists of half the total possible permutations. Indeed, it contains only the cycles of odd length  recall from MATH 133 that flipping two columns will result in flipping the sign of the determinant. So cycles of even length result in a determinant of 1, and cycles of odd length result in a determinant of +1. The terminology gets a bit confusing now, because we refer to a permutation of even length as an "odd" permutation (because the sign is 1), and a permutation of odd length is an "even" permutation (because the sign is +1).
Since we can generalise any cycle as the product of transpositions, this rule makes it easy to compute the sign of any cycle. For example, the 3cycle $(235)$ can be written as $(23)(35)$ (remember that we compose from right to left) and so the determinant is $(1)(1) = +1$.
3Simple groups¶
A group is simple if it has no nontrivial normal subgroups. (Recall that the trivial normal subgroups are $\{e\}$  the group containing only the identity  and $G$  the entire group.) The most basic examples are the groups of prime order, $(\mathbb Z/p\mathbb Z, +)$ (are there others?). These groups have no nontrivial subgroups at all, much less normal subgroups, as a corollary of Lagrange's theorem (since the group order is prime).
3.1Classifying normal subgroups¶
The last question on assignment 5 asked us to classify the normal subgroups of $S_5$. Here's how we actually do this. First we note that any normal subgroup $H \subseteq G$ is a disjoint union of conjugacy classes of $G$ (which are simply sets  specifically, equivalence groups  and not groups). The conjugacy classes in $S_5$ are simply the permutations with cycle decompositions of the same length. So one class contains the 2cycles, another the 3cycles, another the 23 cycles (like $(12)(345)$).
Why is it true that two permutations with samelength cycle decompositions are conjugate? Well, there's probably a really nice proof out there somewhere, but I don't know it. Instead, here's an example that you can use to convince yourself that it's true. Let $g = (123)(45)(67)$, $h = (357)(24)(16)$ (with $g, h \in S_7$). To map $g$ to $h$, we basically need to send $1 \to 3$, $2 \to 5$, $3 \to 7$, etc. Let's create a permutation $a$ that does that: $a = (1376)(254)$. Now, since $a(1) = 3$, then $a^{1}(3) = 1$. So if $g' = aga^{1}$, $g'(3) = 1 \to 2 \to 5$, so $g'(3) = 5$. Also, since $a(2) = 5$, then $a^{1}(5) = 2$. So $g'(5) = 2 \to 3 \to 7$, so $g'(5) = 7$. So far, we have the cycle $(357)$. Starting to look familiar?

He kept saying "sigma" during the lecture but the symbol he drew looked more like delta so let's assume he meant delta. ↩