Group mod a subgroup


by Zorba
Tags: subgroup
Zorba
Zorba is offline
#1
Mar27-11, 12:15 PM
P: 77
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!
Phys.Org News Partner Science news on Phys.org
Going nuts? Turkey looks to pistachios to heat new eco-city
Space-tested fluid flow concept advances infectious disease diagnoses
SpaceX launches supplies to space station (Update)
fourier jr
fourier jr is offline
#2
Mar27-11, 12:22 PM
P: 943
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
micromass
micromass is offline
#3
Mar27-11, 12:27 PM
Mentor
micromass's Avatar
P: 16,583
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).

Deveno
Deveno is offline
#4
Mar29-11, 04:05 AM
Sci Advisor
P: 906

Group mod a subgroup


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).


Register to reply

Related Discussions
Subgroup of a symmetric group Sn Calculus & Beyond Homework 2
Group actions of subgroup of S_3 onto S_3 Calculus & Beyond Homework 2
group action on a subgroup Calculus & Beyond Homework 2
Group/Subgroup Calculus & Beyond Homework 14
Open Subgroup of a Lie Group Differential Geometry 0