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 $g$.

  • $\begingroup$ 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 $(\forall h\in G)(\exists g\in G)[g_i\circ g=h]$ $\endgroup$
    – David P
    May 4 at 1:12
  • 1
    $\begingroup$ If you are asking people to check your argument, please use the [solution-verification] tag. $\endgroup$ May 4 at 2:33
  • 4
    $\begingroup$ Just note that $x\mapsto gx$ Is a bijection on $G$ for every $g\in G$. $\endgroup$
    – citadel
    May 4 at 5:11

1 Answer 1


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 $G$.

  • $\begingroup$ This seems so simple and elegant, I'm forced to ask myself why the other commenters didn't mention it? Am I mistaken in thinking it is simple? $\endgroup$
    – Penelope
    May 4 at 21:06
  • $\begingroup$ @Penelope it is indeed simple. $\endgroup$
    – user264745
    May 4 at 22:55

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