[proofplan]
We prove each assertion directly from the definitions of subgroup, [normal subgroup](/page/Normal%20Subgroup), and [group homomorphism](/page/Group%20Homomorphism). Images of subgroups are checked by writing elements of the image as $\varphi(h)$ and using closure of $H$ under products and inverses. Preimages are checked by applying $\varphi$ to products, inverses, and conjugates, then using the corresponding closure property in $L$. For the final assertion, surjectivity is used exactly to lift an arbitrary conjugating element of $K$ to an element of $G$.
[/proofplan]
[step:Show that the image of a subgroup is closed under the group operations]
Assume $H \le G$. Define the image subset $\varphi(H) \subset K$ by $\varphi(H) := \{\varphi(h) : h \in H\}$.
Since $e_G \in H$ and $\varphi(e_G) = e_K$, we have $e_K \in \varphi(H)$, so $\varphi(H)$ is nonempty. Let $x, y \in \varphi(H)$. Then there exist $a, b \in H$ such that $x = \varphi(a)$ and $y = \varphi(b)$. Since $H \le G$, the product $ab \in H$ and the inverse $a^{-1} \in H$. Therefore
\begin{align*}
xy = \varphi(a)\varphi(b) = \varphi(ab) \in \varphi(H).
\end{align*}
Also,
\begin{align*}
x^{-1} = \varphi(a)^{-1} = \varphi(a^{-1}) \in \varphi(H).
\end{align*}
Thus $\varphi(H)$ is a nonempty subset of $K$ closed under products and inverses, so $\varphi(H) \le K$.
[guided]
Assume $H \le G$. We must prove that $\varphi(H)$ is a subgroup of $K$, so we verify the subgroup conditions directly. The image subset is
\begin{align*}
\varphi(H) := \{\varphi(h) : h \in H\} \subset K.
\end{align*}
First, $\varphi(H)$ is nonempty. Since $H$ is a subgroup of $G$, it contains the identity element $e_G$. Because $\varphi: G \to K$ is a group homomorphism, it sends the identity of $G$ to the identity of $K$, so $\varphi(e_G) = e_K$. Hence $e_K \in \varphi(H)$.
Next, take arbitrary elements $x, y \in \varphi(H)$. By the definition of image, there are elements $a, b \in H$ such that $x = \varphi(a)$ and $y = \varphi(b)$. Because $H$ is a subgroup, it is closed under the product in $G$, so $ab \in H$. Using the homomorphism property of $\varphi$, we get
\begin{align*}
xy = \varphi(a)\varphi(b) = \varphi(ab).
\end{align*}
Since $ab \in H$, the element $\varphi(ab)$ lies in $\varphi(H)$. Therefore $xy \in \varphi(H)$.
Finally, take $x \in \varphi(H)$, so $x = \varphi(a)$ for some $a \in H$. Since $H$ is a subgroup, $a^{-1} \in H$. Since homomorphisms preserve inverses,
\begin{align*}
x^{-1} = \varphi(a)^{-1} = \varphi(a^{-1}).
\end{align*}
Because $a^{-1} \in H$, this shows $x^{-1} \in \varphi(H)$. We have shown that $\varphi(H)$ is nonempty and closed under products and inverses inside $K$, hence $\varphi(H) \le K$.
[/guided]
[/step]
[step:Show that the preimage of a subgroup is closed under the group operations]
Assume $L \le K$. Define the set-theoretic preimage
\begin{align*}
\varphi^{-1}(L) := \{g \in G : \varphi(g) \in L\}.
\end{align*}
Since $\varphi(e_G) = e_K$ and $e_K \in L$, we have $e_G \in \varphi^{-1}(L)$. Let $a, b \in \varphi^{-1}(L)$. Then $\varphi(a), \varphi(b) \in L$. Since $L \le K$, we have $\varphi(a)\varphi(b) \in L$ and $\varphi(a)^{-1} \in L$. By the homomorphism property,
\begin{align*}
\varphi(ab) = \varphi(a)\varphi(b) \in L.
\end{align*}
Thus $ab \in \varphi^{-1}(L)$. Also,
\begin{align*}
\varphi(a^{-1}) = \varphi(a)^{-1} \in L.
\end{align*}
Thus $a^{-1} \in \varphi^{-1}(L)$. Therefore $\varphi^{-1}(L)$ is a subgroup of $G$.
[/step]
[step:Use conjugation in $K$ to prove normality of the preimage]
Assume $L \trianglelefteq K$. By the previous step, $\varphi^{-1}(L) \le G$. To prove normality, let $g \in G$ and let $x \in \varphi^{-1}(L)$. Then $\varphi(x) \in L$. Since $L \trianglelefteq K$ and $\varphi(g) \in K$, the conjugate $\varphi(g)\varphi(x)\varphi(g)^{-1}$ belongs to $L$. Using the homomorphism property,
\begin{align*}
\varphi(gxg^{-1}) = \varphi(g)\varphi(x)\varphi(g^{-1}) = \varphi(g)\varphi(x)\varphi(g)^{-1} \in L.
\end{align*}
Hence $gxg^{-1} \in \varphi^{-1}(L)$. Since this holds for every $g \in G$ and every $x \in \varphi^{-1}(L)$, we have $\varphi^{-1}(L) \trianglelefteq G$.
[/step]
[step:Use surjectivity to lift conjugation in $K$ back to conjugation in $G$]
Assume $\varphi$ is surjective and $N \trianglelefteq G$. By the first step, $\varphi(N) \le K$. It remains to prove normality in $K$.
Let $k \in K$ and let $y \in \varphi(N)$. Since $\varphi$ is surjective, there exists $g \in G$ such that $\varphi(g) = k$. Since $y \in \varphi(N)$, there exists $n \in N$ such that $y = \varphi(n)$. Because $N \trianglelefteq G$, the conjugate $gng^{-1}$ belongs to $N$. Therefore
\begin{align*}
kyk^{-1} = \varphi(g)\varphi(n)\varphi(g)^{-1} = \varphi(gng^{-1}) \in \varphi(N).
\end{align*}
Thus every conjugate in $K$ of every element of $\varphi(N)$ again lies in $\varphi(N)$. Hence $\varphi(N) \trianglelefteq K$.
[/step]
