Does G\H Form a Subgroup of G?

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Discussion Overview

The discussion centers around whether the set difference G\H forms a subgroup of G, particularly in the context of group theory and homomorphisms. Participants explore the implications of normal subgroups and the distinction between set difference and quotient groups.

Discussion Character

  • Homework-related
  • Debate/contested
  • Technical explanation

Main Points Raised

  • One participant suggests that G\H should not be a subgroup since the identity element is in H and thus not in G\H.
  • Another participant states that if H is a normal subgroup, then G/H is a group, but emphasizes that G/H is not a subgroup of G.
  • A different participant clarifies that G\H (set difference) is not a group and therefore cannot be a subgroup, while G/H (quotient group) is a group when H is normal.
  • One participant explains how G/H can be structured as a group through coset multiplication, highlighting the necessity of normality for well-defined multiplication.
  • There is a discussion about the canonical homomorphism from G to G/H and the identity element in the context of quotient groups.

Areas of Agreement / Disagreement

Participants generally agree that G\H is not a subgroup and that G/H is a group when H is normal. However, there is some confusion regarding the notation and the implications of these concepts, indicating a lack of consensus on the initial question.

Contextual Notes

There is a potential misunderstanding regarding the notation G\H versus G/H, which may contribute to the confusion in the discussion. The implications of normality and the definitions of subgroup versus quotient group are also points of contention.

Zorba
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Suppose G is a group and H is a subgroup of G. Then is G\H a subgroup itself? My feeling is it shouldn't be since 1[tex]\in[/tex]H, therefore 1[tex]\notin[/tex]G\H?

I'm getting a bit confused about this because I'm doing a homework sheet and the question deals with a homomorphism from G [tex]\rightarrow[/tex] G\H and I'm wondering what happens to the identity element...

Thanks!
 
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if H is a normal subgroup, then G/H is a group. It's not a subgroup of G though since its elements are the cosets [itex]eH=H, g_1H, g_2H, ...[/itex], which are the images of the elements of G under that mapping. maybe I misunderstood completely though, did you get / mixed up with \ ? because G\H usually means the set {g in G | g not in H}, while G/H means quotient group
 
I think your confusing some notation. G/H is different from G\H.
G\H is the set-difference and it is not a group. And certainly not a subgroup.
G/H is the quotient group, and it is a group (when H is normal). But it's not a subgroup (in general).
 
when H is a normal subgroup of G, G/H can be made into a group by setting coset multiplication as: (Hx)(Hy) = H(xy).

the condition of normality is necessary, because Hx is not uniquely determined by x (it has other members of G in it, as well), and if Hx ≠ xH (equivalently if xHx^-1 ≠ H), then the multiplication will not be well-defined.

there is a "canonical" homomorphism G-->G/H for any quotient group G/H of G, which sends the element g of G to the right (= left) coset Hg. the identity of G/H is He = H, which is also the kernel of the canonical homomorphism.

one way to look at such quotient (or factor) groups is to think of all the members of H being set arbitrarily to the identity, e. since the identity commutes with everything, the normality condition is akin to saying H has to "commute with every g" (this is NOT to say that hg = gh for every h in H. remember, we're "shrinking H to a point" so we can't really tell the difference between h and h' in H. so hg = gh' is the best we can say, they both get packed down into the coset Hg).
 

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