Talk:Quotient group

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Rank of a quotient group

I'm trying to find something to support/refute the idea that

$Rank(N/G)=Rank(N)-Rank(G)$

I can't find it on this page, is it anywhere?

Darcourse (talk) 11:53, 23 January 2022 (UTC)[reply]

Example in which the set of products of two left cosets of a non-normal coset does not even equal a coset

I'm not sure about how useful it would be in the main article, but I found an example using S₃ that shows that the set of products of one of the two left cosets of one of the subgroups of order 2 (all of which are non-normal) with itself consists of four elements:

Let H be the subgroup of S₃ generated by (1 2), meaning H={(),(1 2)}; then the left coset (1 3)H={(1 3),(1 2 3)}, and the product set (1 3)H(1 3)H={(),(1 2),(2 3),(1 3 2)}, which is not a coset of H (or of any other subgroup, because 4 does not divide 6).
The StackExchange answer goes on to show how, if the operation were instead defined as (aH)(bH)=(ab)H, then it would not be well defined: Clearly, (1 3)H=(1 2 3)H, but (1 3)H(1 3)H=H, while (1 2 3)H(1 2 3)H=(1 3 2)H≠H.

As a bit of original research, if N={(),(1 2 3),(1 3 2)} (which is normal because it has index 2), then the other coset is (1 2)N={(1 2),(1 3),(2 3)} and the products of elements of this coset with itself are

(1 2)(1 2)=()
(1 2)(1 3)=(1 3 2)
(1 2)(2 3)=(1 2 3)
(1 3)(1 2)=(1 2 3)
(1 3)(1 3)=()
(1 3)(2 3)=(1 3 2)
(2 3)(1 2)=(1 2 3)
(2 3)(1 3)=(1 3 2)
(2 3)(2 3)=()

from which the product set (1 2)N(1 2)N=N, as expected; I suspect that the product set of two left cosets of a finite non-normal subgroup H, in which the right factor is not H itself, has cardinality equal to |H|² divided by the order of the normal core of H (this relationship obviously holds for normal subgroups, and in that case even if the right factor is the subgroup itself).

Also, the proof in the article that G/H is a group iff H◁G looks a bit like original research, but a very similar proof is given in Dummit & Foote 3ed, p81. Julyo (talk) 04:34, 26 March 2022 (UTC)[reply]

Do we assume that G is a finite group?

In the definitions section, there is both -g H g = H and -g H g included in H but they are not equivalent in infinite groups. Ahsiwje (talk) 05:30, 12 June 2024 (UTC)[reply]