Prove that every row and every column of the multiplication table of a group contains all elements of the group exactly once

abstract-algebragroup-theoryproof-writingsolution-verification

I am self-learning abstract algebra via Aluffi's Chapter 0. I was just curious if someone could proofread this proof and let me know if you think it is correct / rigorous enough (Note that I only proved the rows to keep this post short). Thank you, I would appreciate any feedback.

Exercise: Aluffi II.1.5

The 'multiplication table' of a group is an array compiling the results of all multiplications $gh$:

*  |e   ... h   ...
---+---------------
e  |e   ... h   ...
...|... ... ... ...
g  |g   ... gh  ...
...|... ... ... ...

(Here $e$ is the identity element. Of course the table depends on the order in which the elements are listed in the top row and leftmost column.) Prove that every row and every column of the mulitplication table of a group contains all elements of the group exactly once (like Sudoku diagrams!).

My proof for rows.

Consider a row of group $G$'s multiplication table by fixing a column element $g \in G$. We can use this row to define a set $R_g = \{ gh : h \in G \}$. To prove that every row contains every element of $G$ exactly once, it suffices to show (1) $R_g = G$ and (2) Each entry in the row contains a unique element of $G$.

(1) To prove $R_g = G$, we show that $R_g \subseteq G$ and $Rg \supseteq G$.

( $\subseteq$ ) Let $k \in R_g$. By definition of $R_g$, $k = gh$ for $g, h \in G$. Since $gh \in G$ by closure of groups, it follows that $k \in G$.

( $\supseteq$ ) Let $k \in G$. Every element of $G$ has an inverse, so certainly $g^{-1}k \in G$. Then $g(g^{-1}k) \in R_g$ by definition of $R_g$.
Then $g(g^{-1}k) = (gg^{-1})k = ek = k \in R_g$, where the first equality holds by associativity of elements in $G$.

So $R_g = G$ as desired.

(2) To prove that the row corresponding to $g$ contains all elements of $G$ exactly once, consider two entries $gh$ and $gk$ for $h, k \in G$. Note that $gh = gk \Leftrightarrow h = k$. But by construction of the multiplication table, there is only one entry for each element of $G$. Thus, entries are unique.

With (1) and (2) it follows that each row of $G$'s multiplication table contains all elements of $G$ exactly once. $\square$

Best Answer

This looks OK but you're making a mountain out of a molehill, this is a one-liner.

Suppose a group has distinct elements $a,b,c,d$, and $ab=d$ and $ac=d$ then $b=a^{-1}ab=a^{-1}d=a^{-1}ac=c$. So the multiplication table for $a$ cannot contain duplicates, and hence must contain all of the elements exactly once.

(The condition $ab=d$ and $ac=d$ is equivalent to having two identical elements ($d$) in the multiplication table for $a$. This is a "proof by contradiction".)

Related Question