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

G = \{1, i, -1, -i, C, Ci, -C, -Ci\},

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

S = \{1, -1, C, -C\}

is normal because

Sg = gS

where g is any operation of G. We are given the example g = i

Si = \{i, -i, Ci, iC\} (because iC = -Ci)

iS = \{i, -i, iC, Ci\} (because -iC = Ci)

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

All fine and good.

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

OK, I read that to say that iS and Si of the example on page 251 should form a group, because S is a normal subgroup, and i is a member of G. But iS = \{i, -i, iC, Ci\} is not a group. There is no identity in that set, and i^2 = -1 and -1 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 Sg form a group. So, in our example

\{S, Si, -S, -Si, SC, SCi, -SC, -SCi\}

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

(Sg)(Sh) = S(gh)

I am thinking here that this expression means the members of the two sets Sg and Sh are combined. So in our example with g=1 and h=i

(S)(Si) = \{1, -1, C, -C\}\{i, -i, Ci, iC\}
= \{ i, -i, Ci, iC, -i, i, iC, Ci, Ci, iC, i, -i, iC, Ci, -i, i\}
= \{i, -i, Ci, iC\} = Si

In general S would be the identity

SSg = Sg.

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

-SiSi = -Si^2 = S

And the other group axioms are obeyed.


Actually, our factor group G/S

\{S, Si, -S, -Si, SC, SCi, -SC, -SCi\}

would reduce down to

\{S, Si\}

where

S  = \{1, -1, C, -C\}

and

Si = \{ i, -i, Ci, iC \}

This reduction is possible because all the members of the first set \{S, Si, -S, -Si, SC, SCi, -SC, -SCi\} are equivalent to one of the two members of the second set \{S, Si\}. Like, S = -S because

S  = \{1, -1, C, -C\} = \{-1, 1, -C, C\} = -S.

This is what Penrose means when he says "Note that for a given set Sg, the choice of g is generally not unique; we can have Sg_1 = Sg_2 . . .". This also makes sense in light of Exercise [13.09] where Penrose tells up the order of G/S should be the order of G divided by the order of S, or in our example 8/4 = 2.


17 Aug 2009, 03:49

Joined: 26 Mar 2010, 04:39
Posts: 109
Post Re: 13.2 Question About Normal Subgroups Forming Groups
Yes, you are correct. The cosets Sg, g\in G of the normal subgroup S form a group (where we must identify elements Sg and Sh if g^{-1}h\in S), with identity element given by the coset S and multiplication defined by (Sg)(Sh)=S(gh). 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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