kernel of homomorphisms $C_4 \times C_4 \to C_4$, $V \times V \to V$

Question

The statements are as follows:

1. The group $G = C_4 \times C_4$ has a surjective homomorphism to $C_4$ given by $(g_1, g_2)\to g_1$. The kernel has index 4.
2. The group $G = V \times V$ (where $V$ is the Klein 4 group) has a surjective homomorphism to $V$. The kernel has index 4.

I'm trying to verify these two statements. But I'm confused as to how to find the kernel for these two homomorphisms. The kernel has index 4 means that the order of the kernel is 4. For 1., that means the kernel of $C_4 \times C_4 \to C_4$ is just $C_4$. But elements of $C_4$ cannot be all mapped to the identity simultaneously since if the function is $(g_1, g_2)\to g_1^4$, then the order of kernel should just be 2. (Identity and the element in $C_4$ that has order of 4) For 2., the kernel of $V \times V \to V$ also has order 4. And the statement is implying $V$ is the kernel. Since every element in $V$ has order 2 then if the function is $(g_1, g_2)\to g_1^2$, it makes sense to say $V$ is the kernel. Is my understanding for 2. correct?

Answer 1

For any group $G$ we have a surjective homomorphism $h:G×G\to G$ given by $h(g_1,g_2)=g_1$. Because $$h((g_1,g_2)\cdot (g_3,g_4))=h(g_1g_3,g_2g_4)=g_1g_3=h(g_1,g_2)\cdot h(g_3,g_4)$$.

The kernel, $\{e\}×G$, is isomorphic to $G$.