# proof that every element of finite group G appears once in Cayley table column

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 $$g$$.

• Your last paragraph may need to be more rigorous. You "note that" the column has $n$ elements, which is what you are trying to prove. Perhaps, for the column under $g_i$ and an arbitrary $h\in G$, find an explicit expression for an element $g$ such that $g_i\circ g = h$, and confirm that $g\in G$. This shows $(\forall h\in G)(\exists g\in G)[g_i\circ g=h]$ May 4 at 1:12
• If you are asking people to check your argument, please use the [solution-verification] tag. May 4 at 2:33
• Just note that $x\mapsto gx$ Is a bijection on $G$ for every $g\in G$. May 4 at 5:11

Cayley table of group $$G=\{g_1,…,g_n\}$$ is following: $$\begin{array}{c|ccccc} \circ & g_1 & \cdots & g_j & \cdots & g_n\\\hline g_1 & & & & & \\ \vdots & & & & & \\ g_i & & & g_i\circ g_j & & \\ \vdots & & & & & \\ g_n & & & & & \end{array}$$
Fix $$j$$th column. We claim elements of $$j$$th column are distinct. Assume towards contradiction, $$g_i\circ g_j=g_k\circ g_j$$, for some $$i\neq k$$. By right cancellation law, $$g_i=g_k$$. Thus we reach contradiction. Hence elements of $$j$$th column are distinct. That is, $$j$$th column has $$n$$ distinct elements of group $$G$$. Since order of $$G$$ is $$n$$, every element of $$G$$ occurs exactly once in each column of the cayley table of $$G$$.