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?