# Monday, December 3, 2012 Wrap-up for group theory

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 10-12.

## 2Matrix permutations¶

Last time, we saw that $S_4/V \cong S_3$ where $V$ is the group of 2-2 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 orientation-reversing, 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$(called the alternating group) consists of half the total possible permutations. Indeed, it contains only the cycles of odd length - recall from MATH 133 that flipping two columns will result in flipping the sign of the determinant. So cycles of even length result in a determinant of -1, and cycles of odd length result in a determinant of +1. The terminology gets a bit confusing now, because we refer to a permutation of even length as an "odd" permutation (because the sign is -1), and a permutation of odd length is an "even" permutation (because the sign is +1). Since we can generalise any cycle as the product of transpositions, this rule makes it easy to compute the sign of any cycle. For example, the 3-cycle$(235)$can be written as$(23)(35)$(remember that we compose from right to left) and so the determinant is$(-1)(-1) = +1$. ## 3Simple groups¶ A group is simple if it has no non-trivial normal subgroups. (Recall that the trivial normal subgroups are$\{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, +)$(are there others?). These groups have no non-trivial subgroups at all, much less normal subgroups, as a corollary of Lagrange's theorem (since the group order is prime). ### 3.1Classifying normal subgroups¶ The last question on assignment 5 asked us to classify the normal subgroups of$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$are simply the permutations with cycle decompositions of the same length. So one class contains the 2-cycles, another the 3-cycles, another the 2-3 cycles (like$(12)(345)$). Why is it true that two permutations with same-length cycle decompositions are conjugate? Well, there's probably a really nice proof out there somewhere, but I don't know it. Instead, here's an example that you can use to convince yourself that it's true. Let$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?

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