There is a unique noncyclic group of order 4

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.


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 a previous result, G is abelian.

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