The following exercise is from "Guide to Abstract Algebra, 1st Ed, Carol Whitehead".
Let $(G, \circ)$ be a finite group. Prove that every element of $G$ occurs exactly once in each column of the Cayley table of $G$.
Question: Is my solution below correct?
Note: I ask particularly about the last argument about all elements of $G$ being in the column?
The group $G$ is finite, so we can refer to its elements as $g_1,\; g_2, \; g_3, \ldots g_n$.
Let's consider a column $g$ with elements $ (g_1 \circ g),\; (g_2 \circ g), \; (g_3 \circ g), \ldots (g_n \circ g) $.
Because $G$ is a group, every $(g_i \circ g) \in G$. We conclude that every element of the column $g$ is a member of $G$.
Now consider the equation, $x \circ g = g_i$. For any given $g_i$ there is a unique solution $x$. So if $g_i$ appears in the column $g$, then it can only appear at row $x$. That is, $g_i$ can't appear more than once in the column.
We have not yet shown that every element of $G$ appears in the column $g$. To do this we note that there are $n$ elements in $G$ and therefore there are $n$ elements in the column $g$. Since $g_i$ can't appear more than once in the column $g$, then every element of $G$ appears once in the column $g$.