Algebra Solutions: February 2020

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.

Algebra Solutions

A finite direct product is abelian if and only if each direct factor is abelian

There is a unique noncyclic group of order 4