Assume that is a group of order with identity . Assume also that 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 . Deduce that is abelian.

Let be distinct nonidentity elements of . If , then by left cancellation we have , a contradiction. So is either or the third nonidentity element. Also, if we have since otherwise .

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 . Then , and there are two possibilities for . If , then we have and so , a contradiction. Hence . There are two possibilities for . If , then and by left cancellation we have , a contradiction. Hence . Now since must have an inverse, we have . Now there are three possibilities for . If , we have and so , a contradiction. If , then we have and so , a contradiction. Hence . But now we have , , and , so , a contradiction. Hence .

Now we have . There are two possibilities for . If , then we have so that , a contradiction. Hence . Now there are three possibilities for . If , then , so that , a contradiction. If , then so that , a contradiction. Hence . Now there are two possibilities for . If , we have so that , a contradiction. Hence . Similarly, there are two possibilities for . If we have so that , a contradiction. Hence . There are three possibilities for . If , then we have so that , a contradiction. If , then and so , a contradiction. Thus . There are two possibilities for . If , then so that , a contradiction. Hence . Likewise there are two possibilities for . If we have so that , a contradiction. Hence . Finally, there are three possibilities for . If , we have so that , a contradiction. If , we have so that , a contradiction. Thus .

Thus we have uniquely determined the group table for , 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 is a symmetric matrix. By a previous result, is abelian.