Mathematics 351 – Abstract Algebra 1 Discussion 3 – Alternating groups; Cayley’s Theorem November 26, 2007

Background

Before the break, we saw that the symmetric group S_{n} contains a normal subgroup A_{n} whose elements are all of the “even” permutations, the permutations σ whose corresponding n × n permutation matrices ρ(σ) have determinant 1:

A_{n} = {σ ∈ S_{n} : det(ρ(σ)) = 1}.

A_{n} is called the alternating group on n letters. Today we will study the symmetric and alternating groups and develop some more of their properties.

Discussion Questions

I. A) List all of the permutations in A_{3} and A_{4} using disjoint cycle decomposition notation.

B) Determine the group tables for A_{3} and A_{4}. (Comment: The table for A_{4} will be slightly tedious. You can “parallelize” the computation (split it up among your group members), though. This is an important group we have not seen before, so it is worth the effort!)

C) One fairly common misconception about Lagrange’s Theorem is to think that if k divides |G|, then there must be a subgroup H ⊂ G with |H| = k. But is that true for all of the divisors of |A_{4}|? Are there subgroups of A_{4} of all the possible orders allowed by Lagrange’s Theorem?

D) You should have seen that A_{3} and A_{4} have half the number of elements of S_{3} and S_{4} respectively. Prove in general that |A_{n}| = \frac{1}{2}n!.

II. One of the reasons that the symmetric groups are important examples is a famous result called Cayley’s Theorem. In intuitive terms, this result says that if we understand the symmetric groups and their subgroups, then we understand every finite group G, because every finite group G is isomorphic to a subgroup of S_{n} for some n. In this problem, you will look at some examples and work through one proof of this theorem.

Let G be a finite group of order n and list the elements as

(1)  G = {g_{1}, g_{2}, …, g_{n}}

in some order (the exact order is immaterial – any order will work as long as you maintain it consistently). As we know from a question on the last discussion, each row and column of the group table for G contains each element of G exactly once. If we look at the columns, this means that the left multiplication mappings

\begin{matrix}L_{g} : G & → G \\ x & ↦ gx \end{matrix}

are permutations of G (one-to-one, onto mappings from G to itself). We can identify each one with a permutation of {1, 2, …, n} via the listing of the group elements from (1) above.

A) Let G be a cyclic group of order 4, say with a as generator. Write down the permutation of G defined by L_{g} for each element of G, both in disjoint cycle notation, and as a permutation matrix.

B) Generalize your results from part A to cyclic groups of order n for any integer n ≥ 2.

C) Repeat part A for G = S_{3}. (Be careful! What size will the permutation matrices be?)

Now we want to develop a proof of:

Cayley’s Theorem. Every finite group G is isomorphic to a subgroup of S_{n} for some n.

The n here will actually be n = |G|. We introduce the following notation: Let G̃ be the set of all the left multiplication mappings L_{g} : G → G by elements of G:

G̃ = {L_{g} : g ∈ G},

viewed as permutations in S_{n}.

D) Show that G̃ is a group under the operation of composition of functions. Hence it can be viewed as a subgroup of S_{n} by identifying each L_{g} mapping with its corresponding permutation in S_{n}.

E) Show that the mapping φ : G → G̃ defined by φ(g) = L_{g} is an isomorphism of groups. This completes the proof of Cayley’s Theorem.