Let $G_1$ and $G_2$ be groups and $\pi_1:G_1\times G_2\rightarrow G_1$ Let $G_1$ and $G_2$ be groups and $\pi_1:G_1\times G_2\rightarrow G_1$ be the function defined by $\pi_1(a,b)=a$. Prove that $\pi_1$ is a homomorphism, find $\ker(\pi_1)$, and prove $(G_1\times G_2)/\ker(\pi_1)$ is isomorphic $G_1$.
I've already shown the first two parts of the question, but I'm confused on how to proceed for the last proof.
First, let $G_1$ and $G_2$ be groups. Let $\pi_1:G_1\times G_2\rightarrow G_1$ be the function defined by $\pi_1(a,b)=a$. Let $a_1,a_2\in G_1$ and $b_1,b_2\in G_2$. Then, $\pi_1(a_1a_2,b_1b_2)=a_1a_2$, but $a_1=\pi_1(a_1,b_1)$ and $a_2=\pi_1(a_2,b_2)$, so $\pi_1(a_1a_2,b_1b_2)=\pi_1(a_1,b_1)\pi_1(a_2,b_2)$. $\pi_1$ is a homomorphism. Next, note $\ker(\pi_1) = \{(a,b)\mid \pi_1(a,b)=e_1\}$. Since $G_1$ is a group, $e_1$ is unique, and thus $\ker(\pi_1)=\{(e_1,b)\mid b\in G_2\}$. We can write $\ker(\pi_1)=e_1\times G_2$.
From this point on I think my confusion comes in as to what the structure of $(G_1\times G_2)/\ker(\pi_1)$ is? I don't exactly know what this looks like or how to proceed in proving isomorphic to $G_1$. Is there a simple way to go about this? I think I do have to show some arbitrary map $\phi$ is a bijection from $(G_1\times G_2)/\ker(\pi_1)$ to $G_1$?
Edit: We do not yet have the First Isomorphism Theorem. We have proven the Fundamental Homomorphism Theorem-- which is often called the same thing, but is stated in a different way.

All you need to know is the first isomorphism theorem. This map is a surjection onto $G_{1}$.
So $(G_{1}\times G_{2})/\ker(\pi_{1})\cong G_{1}$.
If you want to proceed directly by hand then you can define $f:(G_{1}\times G_{2})/\ker(\pi_{1})\to G_{1}$ by $f((a,b)\ker(\pi_{1}))=a$. Now observe that the $\ker(f)=\{(e,b)\in G_{1}\times G_{2}:b\in G_{2}\}=\ker(\pi_{1})$. which is the identity element in $G_{1}\times G_{2}/\ker(\pi_{1})$. This gives you injectivity. And for any $a\in G_{1}$ . The equivalence class $(a,e)\ker(\pi_{1})\in (G_{1}\times G_{2})/\ker(\pi_{1})$ is the preimage. Hence done.
Note that the equivalence classes can be written as $(a,G_{2})\ker(\pi_{1})$ as only the first components give different equivalence class. $(a,b_{1})\ker(\pi_{1})=(a,b_{2})\ker(\pi_{1})\,\,,\forall b_{1},b_{2}\in G_{2}$. this is because $(a,b_{1})^{-1}\circ (a,b_{2})=(e,b_{1}^{-1}b_{2})\in \ker(\pi_{1})$.
