Does G/H Form a Subgroup of G?

In summary, G/H is a quotient group and not a subgroup of G. It is defined when H is a normal subgroup of G and is created by setting the coset multiplication as (Hx)(Hy) = H(xy). The condition of normality is necessary for this multiplication to be well-defined. There is a canonical homomorphism G-->G/H which sends the element g of G to the right coset Hg. The identity of G/H is He = H, which is also the kernel of the canonical homomorphism. G/H can be thought of as all the members of H being set to the identity element e, and the normality condition is akin to saying H has to "commute with every g."
  • #1
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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  • #2
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
 
  • #3
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).
 
  • #4
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).
 
  • #5


I would like to clarify that the statement "1\inH, therefore 1\notinG\H" is incorrect. In fact, 1 is an element of both H and G\H. The notation G\H does not mean "G divided by H," but rather it represents the set of all elements in G that are not in H. Therefore, 1 is not excluded from G\H, but rather it is included in both G and G\H.

To answer the question, yes, G\H forms a subgroup of G. This is because it satisfies the three criteria for being a subgroup: closure, identity, and inverses. Closure means that when you combine two elements in the subgroup, the result is still in the subgroup. In this case, if we take two elements from G\H, the result will still be in G\H since all elements in G\H are not in H. The identity element, 1, is still present in G\H, so the subgroup satisfies the identity criterion. Lastly, for every element in G\H, its inverse is also in G\H since the inverse of an element in G\H is simply the inverse of that element in G, which is still in G\H. Therefore, G\H forms a subgroup of G.

As for the homomorphism from G to G\H, the identity element, 1, is still preserved since it is included in both G and G\H. The homomorphism simply maps elements from G to their corresponding elements in G\H, while keeping the group structure intact. So, there is no need to worry about the identity element in this case.

I hope this clarifies any confusion and helps with your homework. Remember to always double check your notation and definitions to avoid any misunderstandings. Good luck!
 

1. What is a subgroup?

A subgroup is a subset of a group that has all the same properties and operations as the original group. In other words, it is a smaller group that is contained within a larger group.

2. How do we determine if G/H is a subgroup of G?

To determine if G/H is a subgroup of G, we must check if G/H satisfies the three conditions for a subgroup: closure, identity element, and inverses. This means that for any elements a and b in G/H, their product must also be in G/H, the identity element of G must also be in G/H, and every element in G/H must have an inverse in G/H.

3. Can G/H be a subgroup if H is not a subset of G?

No, in order for G/H to be a subgroup, H must be a subset of G. This is because all elements of H must also be elements of G in order for the subgroup to inherit the same properties and operations as the original group.

4. What is the significance of G/H being a subgroup of G?

If G/H is a subgroup of G, it means that the elements in G/H can be used to represent cosets of H in G. This has important implications in group theory, as it allows us to study the structure of G by examining the cosets of H.

5. Can G/H be a subgroup if G is not a group?

No, G/H can only be a subgroup if G is a group. This is because the properties and operations of a group are necessary for G/H to also be a group, and if G is not a group, then G/H cannot satisfy these conditions.

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