Is $\ker(\operatorname{nat}_H)=H$? This question came in the exam today, sadly I couldn't answer it.
The question said:

Prove whether or not this is a true statement, stating the reason.
  $$\ker(\operatorname{nat}_H)=H$$
  where $\operatorname{nat}_H$ is natural homomorphism.*

I have a limited knowledge on group theory, so I couldn't answer this question, also I don't think the question provided enough information.

* (From the comments) $H \triangleleft G$ is a normal subgroup of $G$ and $\operatorname{nat}_H : G \to G/H$ is the canonical homomorphism $g \mapsto gH$.
 A: In order for $\operatorname{nat}_H$ to be well defined as a homomorphism, we must assume that $H$ is a normal subgroup (of, let's say, $G$).
Now, given that $H$ is a normal subgroup of $G$, you are meant to prove that $\operatorname{nat}_H(g) = gH = H$ ($H$ the identity element of $G/H$) if and only if $g$ is an element of $H$.
One direction of implication is clear: if $g \in H$, then because $H$ is a subgroup, $\operatorname{nat}_H(g) = gH = H$.  Can you show that this is the case?
For the other direction: if $gH = H$, then for every element $h \in H$: $gh \in H$.  However, because $H$ is a subgroup, $H$ contains the identity element (call it $e$) of $G$.  So, $ge = g \in H$.  So, $g$ is an element of $H$.
A: $G/H$ is a partitioning of $G$ into cosets of $H$. As $H$ is a subgroup of $G$, the image of every $h \in H$ is in the equivalence class of $eH = H$, as $hh' \in H$ for all $h' \in H$ (because $H$ is a subgroup). 
On the other hand, no other $g \in H - G$ is, because if it were then $gH = H$ or $gh' = h''$ for some $h', h'' \in H$, or $g = h''h'^{-1} \in H$. Contradiction. 
Therefore $H$ is the entire kernel.
A: $F(g)=gH$ where $F:G\to G/H$ right?
Claim: $gH=H$ iff $g$ is an element of $H$.
Proof: Suppose $gH=H$ the $g\cdot e$ is an element of $gH$ and belongs to $H$. so $g\cdot e=g$ is an element of $H$.
For the other direction say $g=h_1$ where $h_1$ is an element of $H$ so if $x$ is an element of $gH$ then $x=gh_2=h_1\cdot h_2$ where $h_2$ is an element of $H$. so the product of $h_1$ and $h_2$ belongs to $H$ since $H$ is a group. 
If $x$ belongs to $H$, $x=h_2=h_1\cdot h_1^{-1}\cdot h_2=g\cdot h_1^{-1}\cdot h_2$ so $x$ belong to $gH$. Summing the results, $gH=H$.
So $H$ is the identity in the quotient group $G/H$. 
To find kernel, we solve $F(g)=gH=H$ so kernel is $H$ by the above claim.
A: This is, in essence, a variant of the Fundamental Homomorphism Theorem. So any proof of it runs along the same lines of that theorem.
Note $H \subseteq \text{ker}(\text{nat}_H)$, since if $h \in H, \text{nat}_H(h) = hH = H = e_{G/H}$ (since $H$ is a subgroup, and closed under multiplication- that is left-multiplication by $h$ is a bijection from $H \to H$).
On the other hand, if $g \in \text{ker}(\text{nat}_H)$, then $gH =e_{G/H} =  H$, whence for any $h \in H$, we have $gh \in H$, and thus $g \in Hh^{-1} = H$ (see above). So $\text{ker}(\text{nat}_H) \subseteq H$, and the two are equal.
