# Group mod a subgroup

by Zorba
Tags: subgroup
 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$$\in$$H, therefore 1$$\notin$$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 $$\rightarrow$$ G\H and I'm wondering what happens to the identity element... Thanks!
 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 $eH=H, g_1H, g_2H, ...$, 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
 PF Patron Sci Advisor Thanks Emeritus P: 15,673 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).
P: 905

## 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).

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