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

.