Let $G$ and $H$ be groups.
Prove that $G \times \{1\}$ is a normal subgroup of $G\times H$.
Prove that $(G\times H)/(G \times \{1\})$ is isomorphic to $H$.
My attempt below:
Let $(g_1,1)\in G\times \{1\}$ and $(g_2,1)\in G\times \{1\}$ for some $g_1,g_2\in G$. Since $G$ is a group, it follows from closure that $g_1g_2\in G$. Therefore, $(g_1g_2,1)\in G\times \{1\}$. From the group axioms, $g^{-1}\in G$, which means that $(g^{-1},1)\in G\times \{1\}$, so every element in $G\times \{1\}$ has an inverse. Additionally, $(g^{-1},1)(g,1)=(1,1)\in G\times\{1\}$, so $G\times\{1\}$ and $G\times H$ share the same identity element, which means that $G\times \{1\}$ is a subgroup of $G\times H$. Next, we want to show that $G\times \{1\}$ is normal to $G\times H$. To do this, we must show that $(g',1)(g,h)((g',1))^{-1}$ for some fixed $g'\in G$. So we have that $(g',1)(g,h)((g',1))^{-1}=(g,1)(g,h)(g'^{-1},1)=(g'gg'^{-1},h)$. Since $g'gg'^{-1}\in G$, we have that $(g'gg'^{-1},h)\in G\times H$. Therefore, $G\times \{1\}$ is a normal subgroup of $G\times H$.
I am confused about the last part of my proof, because I assumed some fixed g' and took $(g',1)(g,h)(g'^{-1},1)$ instead of $(g,h)(g',1)(g^{-1},h^{-1})$, although some of these ideas are still new to me, so I'm trying to figure out what mistakes to avoid. Also, I would like some pointers on some assumptions I made earlier that might have ruined the whole proof.
The next part of the proof is to prove that $(G\times H)/(G\times\{1\})$ is isomorphic to $H$. My approach for this one is to use the first isomorphism theorem. That is, to show that there is a map between these two groups that is a surjective homomorphism.
If you know about group morphisms, you can avoid much work!
Consider the group morphism
$$\varphi: G \times H \to H: (g,h) \mapsto h$$
Clearly, this is a surjection with $\operatorname{ker}\varphi = G \times \{1_H\}$. Since the kernel of a group morphism is always a normal subgroup of the domain, we get $ G \times \{1_H\} \unlhd G \times H$ and by the first isomorphism theorem we get
$$(G \times H)/(G \times \{1_H\}) \cong H$$
as desired.