Coset representatives form a group…sometimes?

I am reading “Lie Groups and Lie Algebras and some of Their Applications” by Robert Gilmore and I have become hopelessly confused over what I am sure is a simple matter. With $$H$$ a normal subgroup of the group $$G$$ Gilmore writes:

“A right coset is the set of group operations $$c_0$$, $$c_1$$, ... $$\in G$$ with the property that $$H c_0 + H c_1 + ... = G$$ and, furthermore, no element $$g$$ of $$G$$ is contained more than once in the sum on the left.”

First, is this a non-standard definition of a “coset”? I recall learning that a coset (not “the coset”) was a set of the form $$Hg$$ and that what Gilmore is describing is called a “transversal” of the cosets or a collection of “coset representatives.” He goes on to provide the following theorem:

“If $$H$$ is an invariant subgroup of $$G$$ then the coset elements $$c_0$$, $$c_1$$, $$c_2$$, ... can be chosen in such a way that they are closed under multiplication and form a group called the factor group $$G/H$$

So it seems to me that under Gilmore’s definition we speak of “the coset” of a group $$G$$ with normal subgroup $$H$$ instead of “the cosets” which, the way I recall learning it, are several sets that partition $$G$$ and which are each elements of the factor group. Furthermore I tried to find a multiplicatively closed collection $$c_i$$ for $$D_4$$ (dihedral group of order 8 - the symmetries of a square) with $$H=\{I, R_{180}\}$$ and I can’t do it. I was able to do it for $$A_4$$ with $$H=\{(1), (12)(34), (13)(24), (14)(23)\}$$. Am I just confusing myself somehow?

Gilmore later asks the reader to prove that a collection of coset representatives can always be chosen in such a way that they form the same group as $$G/H$$ when $$H$$ is a normal subgroup of $$G$$. Can someone give me a hint about how to do that proof and maybe demonstrate this for $$D_4$$?

• His treatment sounds eccentric, to say the least. – Lord Shark the Unknown Feb 23 at 17:29
• I couldn't believe it so I checked, but this is really written. Use another book, there are many serious ones. – YCor Feb 23 at 19:24
• So....am I right that this is not possible for $D_4$? Is the theorem even correct? Is it correct always or just under certain circumstances. Isn’t the theorem equivalent to, “We can always find a transversal associated with the cosets of a normal subgroup such that the transversal is itself a group?” – BoundaryValue Feb 23 at 19:45
• I think you are right. It would follow that every quotient of a group is isomorphic to a subgroup. – Berci Feb 23 at 19:53
• Thank you YCor and Berci for your comments. I think I just need to accept that the book has a pretty basic error. I have answered my own question (see answer) and noted that Gilmore actually said what Berci said in a second book! – BoundaryValue Feb 25 at 15:01

“In general it is not possible to chose the group elements of G/H in such a way that they from a subgroup of G. However, if H is an invariant subgroup of G then it is always possible to choose the group elements in the quotient space G/H such that they form a subgroup in G. This group is called the factor group also denoted G/H. Since $$A_3$$ is an invariant subgroup of $$S_3$$ The coset $$S_3 / A_3$$ is a group and this group is isomorphic to $$S_2$$.”