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13.2 Question About Normal Subgroups Forming Groups
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Joined: 07 May 2009, 16:45
Posts: 62
13.2 Question About Normal Subgroups Forming Groups
The symmetry group of the square on page 248 is given as

,

where stands for taking the complex conjugate, we are told on page 251 that the subgroup

is normal because

where is any operation of . We are given the example

(because )

(because )

and since we are not concerned with order of the set members .

All fine and good.

But then we are told on page 252, "Suppose that is a normal subgroup of . Then the distinct sets of , where runs through all the elements of , turn out themselves to form a group." Also "The sets of the form . . . are called cosets of ; but when is normal, the cosets form a group."
(Actually it says ''. . . but when . . ." in my book, but should be as per the correction website)

OK, I read that to say that and of the example on page 251 should form a group, because is a normal subgroup, and is a member of . But is not a group. There is no identity in that set, and and is not in the set. So, that is not a group. What the heck is Penrose talking about?

But on further consideration I think he is saying the set of all the sets form a group. So, in our example

is a group, which is called a factor group. It is a group because,

I am thinking here that this expression means the members of the two sets and are combined. So in our example with and

In general would be the identity

.

And every member would have its inverse. In our example for instance

And the other group axioms are obeyed.

Actually, our factor group

would reduce down to

where

and

This reduction is possible because all the members of the first set are equivalent to one of the two members of the second set . Like, because

.

This is what Penrose means when he says "Note that for a given set , the choice of g is generally not unique; we can have . . .". This also makes sense in light of Exercise [13.09] where Penrose tells up the order of should be the order of divided by the order of , or in our example .

17 Aug 2009, 03:49

Joined: 26 Mar 2010, 04:39
Posts: 109
Re: 13.2 Question About Normal Subgroups Forming Groups
Yes, you are correct. The cosets , of the normal subgroup form a group (where we must identify elements and if ), with identity element given by the coset and multiplication defined by . Note that this multiplication operation is only well defined for normal subgroups.

The elements of the group are sets and we have defined a special multiplication between these sets which provides the group structure.

16 Apr 2010, 04:21
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