# Thread: Show 1-1 and onto between ker(phi) and phi(inverse){(a')}

1. ## Show 1-1 and onto between ker(phi) and phi(inverse){(a')}

OK this question was on a hw assignment i had and noone in the class could answer it and the teacher wouldnt give us the answer so its been bugging me. I think it is kinda like Lagrange's Theorem based on the conclusion but have no idea how to prove it i think the notation and inverses are screwing me up.

Let phi: G---->G' be a homomorphism of groups. Suppose a' = phi(a) element of Range(phi) Let f: ker(phi)--->G be defined as f(k) = ak. Show that f is a 1-1 onto correspondence between ker(phi) and phi(inverse)({a')}. Thus every element of phi(G) has the same sinze inverse image implying that |phi(G)||ker(phi)|=|G|

The only hint he gave was to show a 1-1 for f and for onto to show f(ker(phi)) = phi(inverse){a'}

I really dont have any work to show because i am so confused like i said all the inverses and stuff are throwing me off. I know everything in the kernel is mapped to e' and all a' go back to phi(inverse) of a' but that is all i got. Any help is greatly appreciated

2. Originally Posted by ChrisBickle
OK this question was on a hw assignment i had and noone in the class could answer it and the teacher wouldnt give us the answer so its been bugging me. I think it is kinda like Lagrange's Theorem based on the conclusion but have no idea how to prove it i think the notation and inverses are screwing me up.

Let phi: G---->G' be a homomorphism of groups. Suppose a' = phi(a) element of Range(phi) Let f: ker(phi)--->G be defined as f(k) = ak. Show that f is a 1-1 onto correspondence between ker(phi) and phi(inverse)({a')}. Thus every element of phi(G) has the same sinze inverse image implying that |phi(G)||ker(phi)|=|G|

The only hint he gave was to show a 1-1 for f and for onto to show f(ker(phi)) = phi(inverse){a'}

I really dont have any work to show because i am so confused like i said all the inverses and stuff are throwing me off. I know everything in the kernel is mapped to e' and all a' go back to phi(inverse) of a' but that is all i got. Any help is greatly appreciated
let $k \in \ker \varphi.$ then $\varphi(f(k))=\varphi(ak)=\varphi(a)\varphi(k)=\va rphi(a)=a'.$ so $f(k) \in \varphi^{-1}(a').$ also if $k' \in \ker \varphi,$ then $f(k)=f(k')$ iff $ak=ak'$ iff $k=k'.$ thus $f$ is one-to-one.

to prove that $f$ is onto, let $b \in \varphi^{-1}(a').$ then $\varphi(b)=a'=\varphi(a)$ and hence $\varphi (a^{-1}b)=1,$ which means $a^{-1}b \in \ker \varphi.$ let $a^{-1}b=k.$ then $b=ak=f(k)$ and you're done.

3. It all seems so easy once you see a solution...but thats math isnt it?
And where can i learn to use the actual math symbols its quite annoying trying to write everything out in a form that makes sense

4. Originally Posted by ChrisBickle
It all seems so easy once you see a solution...but thats math isnt it?
And where can i learn to use the actual math symbols its quite annoying trying to write everything out in a form that makes sense
Here to start.

Yeah, I'll be honest with you. I took one look at that and was like "I am not spending the time to decipher the math from the plain text". LaTeX sounds like a good idea