Skip to content

A Group Theory Question

February 19, 2013

Question: What is the minimum number of elements two normal subgroups G* and G’ of G may have such that |G*| = |G’| (where | | denotes the order, or number of elements, of a group) but G* != G’?

A few preliminary definitions for the uninitiated:
def Group: (X, *) where X is a set of elements and * is an operation such that for every two elements x and y in X, x * y is also in X. Additionally, there are a few other conditions:

Associativity: For every three elements x, y, z in X, (x * y) * z = x * (y * z). This holds for most standard operations on most “standard” sets X one is used to thinking about; for example, if X is the set of all integers with the operation of addition, we know associativity of the integers with addition holds (a fact one usually learns in elementary school).

Existence of Identity: That is, there exists some element e in X such that for every element x in X, e * x = x * e = x. Again, using the example of (Z, +) where Z denotes the set of integers, the identity element is zero; one may take any element of Z and add zero to get itself.

Existence of Inverse: That is, for every element x in X there is another element x’ in X such that x * x’ = e. Let us use the same example of (Z, +) — if we take any element a in Z, if we add a + (-a), since -a is also an element of Z, we get the identity element (which is 0 as we found above) in the case of this group (Z, +).

def Subgroup: A subgroup G* = (X’, *) of a group G = (X, *) is a set X’ which is a subset of X (that is, X’ is contained in X but does not equal X) with the identical operation * (of G) such that (X’, *) forms a group.

Example: Take the real numbers with the operation of addition. The integers are a subset of the real numbers, and with addition they also form a group. It follows that (Z, +) is a subgroup of (R, +) where R denotes the set of real numbers.

def Normal Subgroup: A normal subgroup is a subgroup G* of a group G such that for every element x in G x * G* = G* * x, that is for every element in G* we get the same subgroup if, for each y in G*, when we take the set of all x * y it is the same subgroup as if we take the set of all y * x.

Example: Take the dihedral group of four elements D_2 = {1, -1, k, -k} with operation multiplication (with k * k = -1) describing the actions of the identity (1), reflection about the y-axis (-1), and rotation by +- 90 degrees (k and -k respectively) of a square with corners on the four major axes on the typical Cartesian plane (think of a square with sides at y = 1, y = -1, x = 1, and x = -1 rotated 45 degrees).

Exercise: Show that D_2 is a group using the definition of a group provided above

Question: What are the normal subgroups of D_2?

Another way of asking this question is: what subgroups of D_2 can be acted upon by each element of D_2 such that for every element x of D_2 and y of a subgroup does xy = yx?

It seems a complicated question! Let us simplify things a bit — to begin, what are the subgroups of D_2?

It seems rather obvious that {1} is a subgroup. Are there any others?

Exercise: Find all subgroups of D_2 and show they are subgroups by the definition given above

Solution to exercise:
{1}, {1, -1}, {1, -k} are all subgroups of D_2. Which of these subgroups are normal?

Interestingly, it turns out both {1, -1} and {1, -k} are normal subgroups of D_2 (show this as an exercise) and yet they are not identical! There exist two subgroups of D_2 that are normal with order (number of elements) two yet they do not contain the same elements! Since it is assumed that neither G* nor G’ have only one element (otherwise they would be identical which is a contradiction to the statement of the problem) we have stumbled across a simple solution to what seems a rather complicated question at first glance.

About these ads

From → Uncategorized

Leave a Comment

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Follow

Get every new post delivered to your Inbox.

%d bloggers like this: