Prove that $f(g_j) = (e_1, e_2, . . . , g_i, e_(i+1), . . . e_n)$ is a group homomorphism for each i and find the kernel and the image of the function

Question

This is the task im trying to solve:

Can someone please clarify this for me? I don't even know where to begin. If we are mapping from $G_i$, then what the hell is $g_j$ ? I assume an element of $G_i$. Then what is $g_i$ supposed to be? How are we even counting the index i?

I'm really confused on this.

Answer 1

Surely, it is a typo. The correct formula for $\phi_i$ should be $$\phi_i(g) = (e_1,\dots,e_{i-1},g,e_{i+1},\dots,e_n),$$ a kind of inclusion of each $G_i$ into the product $G_1 \times \cdots \times G_n$.