What's the relationship of $e_G$ to $G \times G$?

Question

Context: 1st year Mathematics BSc.

Let \begin{align} & G = \langle g \rangle \text{ be a cyclic group of order } n, \tag{0.1}\\ & e_1 \text{ be the neutral element of } G, \tag{0.2}\\ & e_2 \text{ be the neutral element of } G \times G. \tag{0.3} \end{align} Then \begin{align} e_2=(g^i,g^j) \text{, for some } 1\leq i,j, \leq n, \tag{1} \end{align} so, for all $k \in \mathbb{Z}$, \begin{align} e_2 &=e_2^k \tag{2.1}\\ &=(g^i,g^j)^k \tag{2.2} \\ &=\left((g^i)^k,(g^j)^k\right) \tag{2.3} \\ &=\left(g^{ik},g^{jk}\right), \tag{2.4} \end{align} so, for all $k \in \mathbb{Z}$, \begin{align} & g^i=g^{ik}, \tag{3.1}\\ & g^j=g^{jk}, \tag{3.2} \end{align} so \begin{align} & g=g^k. \tag{4}\\ \end{align}

This kind of looks like it might mean that $$g^i=g^j=e_1, \tag{5}$$

but I don't know if that's right or how to prove it.

Answer 1

The relationship is that $e_2=(e_1,e_1)$. You can check it by doing the multiplication with any other element of the product.

This is true more generally, not just for cyclic $G$.

Answer 2

The neutral element of $G \times G$, call it $e_2$ is exactly $(e_1,e_1)$ where $e_1$ is the identity of $G$. So your conclusion that $g^i = g^j = e_1$ is correct based on the assumption that $e_2 = (g^i,g^j)$. To prove that $e_2 = (e_1,e_1)$ all you have to do is go through the definition of what the identity element of a group must satisfy. Does $(g^k,g^l)(e_1,e_1) = (e_1,e_1)(g^k,g^l) = (g^k,g^l)$ for any choice of $g^k,g^l$? The fact that $G$ is cyclic is not relevant for this reasoning to hold.