Abstract Algebra: need a review of 1-1 and onto proof

[Halaaku]

Homework Statement

define a function f--> gHg$^{-1}$

Homework Equations

prove if f is 1-1 and onto.

The Attempt at a Solution

1-1:
f(h1)=f(h2)
gh1g$^{-1}$=gh2g$^{-1}$
h1=h2 (left and right cancellations)

onto:
f(g$^{-1}$hg)=gg$^{-1}$hgg$^{-1}$=h
so every h belonging to H has an image of g$^{-1}$hg.
However,I do not really understand the last line. I followed the example that we did in class but now I am not sure.

 

[pasmith]

Homework Statement

define a function f--> gHg$^{-1}$

If you tell me what $H$ and $g$ are (I'm assuming there's a group $G$, that $H$ is a subgroup of $G$, and that $g \in G$) I will know the domain and codomain of $f$, but you haven't said anything about what $f(h)$ is for $h \in H$, so I can't say whether $f$ is injective or surjective.

Homework Equations

prove if f is 1-1 and onto.

The Attempt at a Solution

1-1:
f(h1)=f(h2)
gh1g$^{-1}$=gh2g$^{-1}$
h1=h2 (left and right cancellations)

That's one mystery solved: $f: h \mapsto ghg^{-1}$ and it is indeed 1-1.

onto:
f(g$^{-1}$hg)=gg$^{-1}$hgg$^{-1}$=h
so every h belonging to H has an image of g$^{-1}$hg.

No. What you've written here is that if $g^{-1}hg \in H$ then $f(g^{-1}hg) = h \in gHg^{-1}$. But unless $H$ is normal, and you haven't told me that it is, it may not be the case that if $h \in H$ then $g^{-1}hg \in H$.

To prove that $f$ is onto, you need to start with a $k \in gHg^{-1}$ and show that there exists $h \in H$ such that $f(h) = k$.