Show that $N=g_1 + g_2 + \cdots g_n$ is in the center of the group ring $RG$

Question

There is a similar question here which I can understand but there is no mention of conjugacy class in my question so I'm somewhat confused.

Let $G=\{g_1, g_2, ..., g_n \}$ be a finite group, show that $N= g_1 + g_2 + \cdots +g_n$ is in the center of the group ring $RG$.

I tried multiplying out $(a_1g_1 + a_2g_2 + \cdots + a_ng_n)(g_1 + g_2 + \cdots + g_n)$ and $(g_1 + g_2 + \cdots + g_n)(a_1g_1 + a_2g_2 + \cdots + a_ng_n)$ and checking for equality.

Does it work out because $(a_1g_1 + a_2g_2 + \cdots + a_ng_n)(g_1 + g_2 + \cdots + g_n)= (a_1g_1)(g_1 + g_2 + \cdots + g_n)+(a_2g_2)(g_1 + g_2 + \cdots + g_n)+\cdots +(a_ng_n)(g_1 + g_2 + \cdots + g_n) = (g_1 + g_2 + \cdots + g_n)(a_1g_1)+(g_1 + g_2 + \cdots + g_n)(a_2g_2)+\cdots +(g_1 + g_2 + \cdots + g_n)(a_ng_n)= (g_1 + g_2 + \cdots + g_n)(a_1g_1 + a_2g_2 + \cdots + a_ng_n)$

where by definition $(ag_i)(bg_j)=(ab)(g_ig_j)$ and $g_i(g_1 + g_2 + \cdots + g_n)=(g_1 + g_2 + \cdots + g_n)g_i$, multiplying all the elements of $G$ on the left or right by the same element is a permutation of the elements of $G$ thus the sum is unchanged?

Answer 1

The proof you refer to applies here in just the same way.

For any element $g$ of the group, consider $g^{-1}Ng=\sum g^{-1}g_ig$. The elements $g^{-1}g_ig$ are (in some order) precisely all the elements of the group and so $g^{-1}Ng=N$.

N.B. Your comment at the end of your proof is correct. This works because of the 'Latin Square' property of groups.