**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.1 Exercise 1.1.36**

Assume that $G = \{1, a, b, c\}$ is a group of order 4 with identity 1. Assume also that $G$ has no element of order 4. Use the cancellation laws to show that there is a unique (up to a permutation of the rows and columns) group table for $G$. Deduce that $G$ is abelian.

Solution: Let $x,y$ be distinct nonidentity elements of $G$. If $xy = x$, then by left cancellation we have $y = 1$, a contradiction. So $xy$ is either 1 or the third nonidentity element. Also, if $x \neq 1$ we have $x^2 \neq x$ since otherwise $x = 1$. Now we need to find all the possible ways to fill in the following group table under the given constraints.

1 | a | b | c | |
---|---|---|---|---|

1 | 1 | a | b | c |

a | a | |||

b | b | |||

c | c |

Suppose $ab = 1$. Then $ba = 1$, and there are two possibilities for $ac$. If $ac = 1$, then we have $ab = ac$ and so $b = c$, a contradiction. Hence $ac = b$. There are two possibilities for $bc$. If $bc = 1$, then $bc = ba$ and by left cancellation we have $c = a$, a contradiction. Hence $bc = a$. Now since $c$ must have an inverse, we have $c^2 = 1$. Now there are three possibilities for $a^2$. If $a^2 = 1$, we have $a^2 = ab$ and so $a = b$, a contradiction. If $a^2 = b$, then we have $a^2 = ac$ and so $a = c$, a contradiction. Hence $a^2 = c$. But now we have $a^2 = c$, $a^3 = b$, and $a^4 = 1$, so $|a| = 4$, a contradiction. Hence $ab \neq 1$.

Now we have $ab = c$. There are two possibilities for $ba$. If $ba = 1$, then we have $ca = aba = a$ so that $c = 1$, a contradiction. Hence $ba = c$. Now there are three possibilities for $a^2$. If $a^2 = b$, then $a^3 = ab = c$, so that $|a| = 4$, a contradiction. If $a^2 = c$, then $a^2 = ab$ so that $a = b$, a contradiction. Hence $a^2 = 1$. Now there are two possibilities for $ac$. If $ac = 1$, we have $ac = a^2$ so that $a = c$, a contradiction. Hence $ac = b$. Similarly, there are two possibilities for $ca$. If $ca = 1$ we have $ca = a^2$ so that $a = c$, a contradiction. Hence $ca = b$. There are three possibilities for $b^2$. If $b^2 = c$, then we have $b^2 = ab$ so that $a = b$, a contradiction. If $b^2 = a$, then $b^3 = ba = c$ and so $|b| = 4$, a contradiction. Thus $b^2 = 1$. There are two possibilities for $bc$. If $bc = 1$, then $bc = b^2$ so that $b = c$, a contradiction. Hence $bc = a$. Likewise there are two possibilities for $cb$. If $cb = 1$ we have $cb = b^2$ so that $b = c$, a contradiction. Hence $cb = a$. Finally, there are three possibilities for $c^2$. If $c^2 = a$, we have $c^2 = bc$ so that $b = c$, a contradiction. If $c^2 = b$, we have $c^2 = ac$ so that $c = a$, a contradiction. Thus $c^2 = 1$. Thus we have uniquely determined the group table for G, as shown below.

1 | a | b | c | |
---|---|---|---|---|

1 | 1 | a | b | c |

a | a | 1 | c | b |

b | b | c | 1 | a |

c | c | b | a | 1 |

Note that the group table for $G$ is a symmetric matrix. By Exercise 1.1.10, $G$ is abelian.

Another proof to show $G$ is abelian.

Since there is no element of order 4, then the order can only be 1 or 2 or 3 by Exercise 1.1.34.

Suppose there exists an element $x$ such that the order of $x$ is 3. Then $1, x, x^2$ are pairwise distinct. Hence there exists another element $y\in G$ such that $y\ne 1,x, x^2$.

We consider $xy$. It is clear $xy$ can not be $1,x,x^2,y$. So we obtain a new element, a contradiction (more than 4 elements). Thus there are no elements of order 3.

Hence $x^2=1$ for all $x\in G$. Now The group $G$ is abelian can be obtained directly from Exercise 1.1.25.