Ordered Pairs and Fundamental Homomorphism Theorem 
Let $G$ and $H$ be groups. Suppose $J$ is a normal subgroup of $G$ and $K$ is a normal subgroup of $H$. Show that the function $f(x, y) = (Jx, Ky)$ is a homomorphism from $G \times H$ onto $(G/J) \times (H/K)$, and find the kernel of $f$.

So, when checking for $f(ab) = f(a)f(b)$ I wasn't sure if I needed both an $(x, y)$ and an $(a,b)$ in $G$, i.e. two different ordered pairs to check? 
I tried $f(a, b) = (Ja, Kb)$ and $f(x, y) = (Jx, Ky)$, and so $$f(a, b)f(x, y) = (Jax = Kby).$$
But then doesn't $f((a, b)(x, y)) = (Jab, Kxy)\neq(Jax = Kby)$?
Is the kernel of $f$ equal to $$\{x, y \in G: (Jx, Ky) = e\}\;?$$ Or would it be $$\{x, y\in G: f(x, y) = e\}\;?$$ I wasn't sure if the second one is more notationally correct or something.
 A: Hint: Recall that if the group operation on a group $G$ is denoted $\cdot$ and the group operation on a group $H$ is denoted $\ast$, then the group operation $\star$ on $G\times H$ is defined to be
$$(g_1,h_1)\star(g_2,h_2)=(g_1\cdot g_2,h_1\ast h_2).$$
The identity element of $G\times H$ under this operation is $(e_G,e_H)$, where $e_G$ and $e_H$ are the identity elements of $G$ and $H$ respectively.
Show that if $\varphi:G_1\to G_2$ and $\psi:H_1\to H_2$ are group homomorphisms, then the function
$$\rho:(G_1\times H_1)\to(G_2\times H_2),\qquad \rho(a,b)=(\varphi(a),\psi(b))$$
is a group homomorphism. Additionally, prove that
$$\ker(\rho)=\ker(\varphi)\times\ker(\psi)$$
where, by definition,
$$\ker(\rho)=\{(a,b)\in G_1\times H_1:\rho(a,b)=(e_G,e_H)\}$$
and, by definition,
$$\ker(\varphi)\times\ker(\psi)=\{(a,b)\in G_1\times H_1:a\in\ker(\varphi),\,b\in\ker(\psi)\}.$$
Apply these observations with $G_1=G$, $G_2=G/J$, $H_1=H$, and $H_2=H/K$.
A: To check if $f$ is a homomorphism, you need to show that:
$f[(g_1,h_1)\ast(g_2,h_2)] = f(g_1,h_1)\ast f(g_2,h_2)$
or, equivalently, that:
$(J(g_1g_2),K(h_1h_2)) = (Jg_1,Kh_1)\ast (Jg_2,Kh_2)$ in $(G/J) \times (H/K)$.
I would write the kernel as:
$\{(g,h) \in G \times H: (Jg,Kh) = (J,K)\}$
since $(J,K) = (e_{G/J},e_{H/K}) =$ the identity of $(G/J) \times (H/K)$.
