email questions to here:

Question: In the proof of Orbit-Stabilizer given in class, you wrote that there are two ways of counting the elements in G: (1) |G|, or (2) the sum over all x' in Orbit(x) of the number of g in G such that gx = x. Could you explain how we get the expression in (2) again?

Answer: Every element g in G sends x somewhere else in the orbit of x. Let's imagine we enumerate the orbit of x as {x_1,...,x_n}. We can divide the elements of G up into those which send x to x_1, those which send x to x_2, those which send x to x_3, and so on. If g maps to x to x_i, then it doesn't also map x to x_j for an x_j =/= x_i. So this is a genuine division of G into different sets. The order of G is then the sum of the orders of all of these different sets. So

|G| = {#elements of G sending x to x_1} + {#elements of G sending x to x_2} + {...} + {#elements of G sending x to x_n}

because the right hand side is just counting the partition of G. If H is the Stabilizer of x, then we identified all the sets above with left cosets of H, so they all have order |H| = |Stab(x)|, and so

|G| = |Stab(x)| * |Orbit(x)|.

Question: The centralizer of an element is a subgroup but I am having trouble coming up with an argument for why the centralizer of a non-trivial element does not form an abelian subgroup of whatever Group the element is in.

Answer: Let C be the centralizer of g. The problem is that if x commutes with g and y commutes with g, then there is no reason why x has to commute with y. Probably the easiest example to consider is the case when g = e. In this case, the centralizer C is the entire group G, because everything commutes with the identity. But certainly all groups need not be abelian.

Question: I am still a little unclear on how to think about the structure of direct products of groups like Z x Z/2Z or Sn X Z/pZ--how can we intuitively see things like the order of elements, the number of elements and things like that?

Answer: The key thing to keep in mind in direct products A x B is that the groups A and B do not interact at all. An element of G = A x B is just a pair (a,b) of elements a in A and b in B. So the number of elements in G is just the number of such choices, which is |A| times |B|. As to thinking about the orders of elements, here is a related problem. What is the order of (1,2,3,4)(6,7) in S_8? Think about how the 4-cycle with just cyclically permute the elements 1,2,3 and 4, and so g^n will fix these numbers exactly when n is a multiple of 4. Similarly, g^n will permuate 6 and 7, and so g^n will fix these numbers when n is a multiple of 2. Can you deduce from this that |g| = 4? In a direct product, to compute the order of g = (a,b) we need to think of the orders of a and b separately. Suppose that |a|=4 and |b|=2. The product law on direct product groups says that g^n = (a^n,b^n). For g^n to be the identity, we need a^n = e_A and b^n = e_B. So we need n to be a multiple of 4 for a^n to be trivial, and n to be a multiple of 2 for b^n to be trivial, so |g| = 4. In general, the order will have to be a multiple of the order of a and the order of b. But the smallest such multiple will do, so the order is the lowest common multiple of |a| and |b|. Suppose we wanted to compute the number of elements of A x B such that g^4 is trivial. Certainly g^4 = (a,b)^4 = e implies that a^4 and b^4 are both trivial. Hence

(#a in A with a^4 = e_A) x (#b in B with b^4 = e_B) = (#g in A x B with g^4 = e)

But suppose we want to know the number of elements of exact order 4. Then we have to subtract the number of elements of order 2 or of order 1. Equivalently, we may subtract the number of elements g with g^2 = e. Now:

(#a in A with a^2 = e_A) x (#b in B with b^2 = e_B) = (#g in A x B with g^2 = e)

The first computation gave elements of order dividing 4, and the second gives elements of order dividing 2. The elements of exact order 4 are precisely those with order dividing 4 and not dividing 2. And so, subtracting the first term from the second term, we get:

# elements of A x B of order exactly 4 = (#a in A with a^4 = e_A) x (#b in B with b^4 = e_B) - (#a in A with a^2 = e_A) x (#b in B with b^2 = e_B).

This actually becomes a little more complicated when "4" is replaced by "6". In that case, we have to subtract the elements of order dividing 3 and order dividing 2, but that subtracts the identity element twice! So we need to add it back in. (There are more and more terms when 6 is replaced by a number with more distinct prime factors.)

Example: Let A = S_4 and B = S_4. You can check that A has 6 elements of order 4, 9 elements of order 2, and 1 element of order 1. Thus S_4 has 16 elements a with a^4 = e, and it has 10 elements with a^2 = e. Thus

The number of elements of S_4 x S_4 of order 4 = 16 * 16 - 10 * 10 = 256 - 100 = 156.

Question: In the practice problem 38, it asks us to compute a left coset. I was wondering is it true that we can write a group as a union of its left cosets? If so how would we go about finding the left cosets for the G w.r.t. H in 38?

Answer: To compute the left coset gH, you simply need to write down the elements gh for every h in H. The result will be a set of | H | such elements Yes, a group is a union of its (left) closets. But computing that decomposition is just the same as computing each individual coset.

Question: Your question goes here!

Answer : My Answer goes here.

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