f: G/G_{s}------>Os, defined by f(gG_{s})=gs.

O_{s}={s'in S st s'=gs for some g in G}

G_{s}={g in G st gs=s}

I am supposed to show f is well defined, 1-1, and onto.

I know that to bewell defined, if a=b then f(a)=f(b). In this case I think that would mean, if g_{1}G_{s}=g_{2}G_{s}, then g_{1}s=g_{2}s. So I assume that g_{1}G_{s}=g_{2}G_{s}is true. Does that mean that g_{1=}g_{2}? If so, can I then substitute a generic g in for g_{1}and g_{2}?

I know1-1means that if f(a)=f(b), then a=b. For this case, if g_{1}s=g_{2}s, then g_{1}G_{s}=g_{2}G_{s}._{So I assume }g1s=g2s, and show g1Gs=g2G_{s}. Since g1s=g2s, g_{1}=g_{2}? Am I allowed to divide by s? I know I can't multiply by the reciprocal because S isn't a group, it's a set.

I know thatontomeans that for every element in the codomain-y_{, }there exists and element in the domain-x with f(x)=y. So in this case, for every element in O_{s}, the orbit of s, there must exist an element in the domain, the quotient group of G_{s}in G (which is the (left) cosets of G_{s}in G) with f(g'G_{s})=g's? I really don't know what to do here?

Any guidance would be appreciated.