Intuitive Understanding of the First Isomorphism Theorem I've been reading some things about an intuitive understanding of the first isomorphism theorem, and there's just one more part that I do not understand. 
Specifically, the first isomorphism theorem states that for a homomorphism $f: G \to H$, $\ker f$ is a normal subgroup of $G$, and there exists an isomorphism $h: G/\ker f \to f(G)$. The intuitive explanation that I read said that the group looks similar to its image under the homomorphism when divided by a certain subgroup, since certain elements which are mapped to the same under the homomorphism are grouped together. 
What I don't intuitively see is why $\ker f$ is a normal subgroup of $G$, and why the certain subgroup is the $\ker f$ (I think the former will help explain the latter).
I know this question is a little more vague, so thanks in advance for your help.
 A: The map $h : G/\ker f \to f(G)$ is essentially the same map as $f$, except that elements that were mapped to $0$ by $f$ are not taken into consideration anymore. 
You can only do this when your subgroup is normal. As an example, consider the group $D_n = \langle r,s \, | \, r^n = s^2=1, srs = r^{-1} \rangle$. Take $n > 2$ for our example, and consider the two maps 
$$
f_1 : D_n \to \mathbb Z/n \mathbb Z, \quad f_1(r^i s^j) = \overline i, \quad 0 \le i < n, \quad j \in \{0,1\} 
$$
$$
f_2 : D_n \to \mathbb Z/2 \mathbb Z, \quad f_2(r^i s^j) = \overline j, \quad 0 \le i < n, \quad j \in \{0,1\} 
$$
(The $\overline{i}$ means $i$ modulo $n$ ; I use the same notation for $j$ modulo $2$.) The first one is not even a group homomorphism ; we do not have the property that $f_1(s)f_1(r)f_1(s) = f_1(r^{-1})$ (check this ; this is where I use $n > 2$), which should have been true if $f_1$ was a group homomorphism. Intuitively, $f_1$ is an attempt to "mod out $s$". The reason why this attempt fails is explained by the fact that the subgroup $\langle s \rangle \subseteq D_n$ is not normal, i.e. $rsr^{-1} = r^2s \neq s$ when $n > 2$. 
The second one, however, works just fine : it is a group homomorphism since 
$$
f_2(r^{i_1} s^{j_1} r^{i_2} s^{j_2}) = f_2(r^{i_1 - i_2} s^{j_1 + j_2}) = \overline{j_1 + j_2} = \overline{j_1} + \overline{j_2}.
$$ 
This is illustrated by the fact that the subgroup $\langle r \rangle \subset D_n$ is normal, i.e. $r(r^i)r^{-1} = r^i$ and $s(r^i)s^{-1} = r^{-i}$. 
So if we dismiss the information contained in a normal subgroup, we are fine ; this is because if $H \le G$ is a normal subgroup, then a product $gh$ where $g \in G$ and $h \in H$ will satisfy $gh = h'g$ where $h$ and $h'$ are elements of $H$. If you "don't read" $h$ and $h'$ in this equation, then you see $g$ on both sides. This is what being a normal subgroup means ; when we consider the elements of $H$ as behaving like the identity element in $G/H$, then we have a consistent group structure. 
When you understand normal subgroups, you understand the projection map $\pi : G \to G/H$ for a normal subgroup $H$. The generalization $f : G \to K$ with $H = \ker f$ is not a big deal ; you can always replace $K$ by $f(G)$ (since $f(G)$ is a subgroup of $K$) and if you have read the proof of the first isomorphism theorem, you can always replace $f(G)$ by $G/H$, up to isomorphism. So your understanding of normal subgroups and the first isomorphism theorem reduces to understanding the projection map $\pi : G \to G/H$. 
Hope that helps,
A: To answer your first question:
Let $a \in \ker{f}$,  we would then need to show that $gag^{-1} \in \ker{f}$. But $f(gag^{-1}) = f(g) f(a) f( g^{-1}) = f(g)f( g^{-1}) = f(gg^{-1}) = f(e)$
Thus $gag^{-1} \in \ker{f}$. 
I'm not sure that I understand your second question, could you rephrase it? 
