If $H \leq G$ and $\ker (f) \subseteq H$, then $f^{-1} (f(H)) = H$.
This is as far as I got:
$H \leq G \implies \forall a,b \in H, ab^{-1} \in H.$
ker($f$) $\subseteq H$ equivalently states $f(x) = e \implies x \in H$
How do I get to show that $f^{-1} (f(H)) = H$? I assume I want to show $\forall x \in H, f^{-1} (f(x)) = x$, but I'm not sure how to do it. Some help would be great.
On
If $h\in H$, then $f(h)\in f(H)\implies h\in f^{-1}(f(H))$.
If $g\in f^{-1}(f(H))$, then $f(g)\in f(H)\implies \exists h$ such that $f(g)=f(h)\implies f(gh^{-1})=e\implies gh^{-1}\in\rm{ker}f\implies gh^{-1}\in H\implies g\in H$.
On
Obviously $H\subset f^{-1}(f(H))$.
On the other hand let $y\in f^{-1}(f(H))$, then there exists $x\in H$ such that $f(y)=f(x)$. Since $f$ is a group homomorphism, $yx^{-1}\in\mbox{ker}(f)\subseteq H$, in particular $y\in H$.
On
If $f\colon X\to Y$ is any map and $A\subseteq X$, then $A\subseteq f^{-1}(f(A))$ holds.
So in your case the interesting bit is the reverse inclusion.
Start with $x\in f^{-1}(f(H))$. This means that $f(x)\in f(H)$, so there exists $y\in H$ such that $f(x)=f(y)$.
You cannot conclude that $x=y$, and this is why the inclusion mentioned at the beginning doesn't generally hold (it does for every subset of the domain if and only if the map is injective).
However, we have some additional assumptions: the map is a homomorphism and $\ker f\subseteq H$. It's a general fact for group homomorphisms that $f(x)=f(y)$ if and only if $xy^{-1}\in\ker f$ (you may want to review it). So in your case $xy^{-1}=z\in\ker f$, but then $$ x=zy\in H $$ because $z\in\ker f\subseteq H$.
Clearly $H \subset f^{-1}(f(H))$.
Conversely, suppose $g \in f^{-1}(f(H))$. Then $f(g) \in H$, so there exists $h \in H$ with $f(g) = f(h)$. It follows that $f(gh^{-1}) = 1$, so $gh^{-1}$ is in the kernel of $f$, so in particular is in $H$.
Then $g = (gh^{-1})h$ is the product of two things in $H$, so it is in $H$ itself.