Hw4 Problem 8

### Problem 8

This exercise shows that there are two nonisomorphic group structures on a set of 4 elements.

Let the set be $\{e,a,b,c\}$, with $e$ the identity element for the group operation. A group table would then have to start in the manner shows in Table 4.22 of the book. The square indicated by the question mark cannot be filled in with $a$. It must be filled in either with the identity element $e$ or with an element different from both $e$ and $a$. In this latter case, it is no loss of generality to assume that this element is $b$. If this square is filled in with $e$, the table can then be completed in two ways to give a group. Find these two tables. If this square is is filled in with $b$, then the table can only be completed in one way to give a group. Find this table. Of the three tables you now have, two give isomorphic groups. Determine which two tables these are, and give the one-to-one onto renaming function which is an isomorphism.

(a) Are all groups of $4$ elements commutative?

(b) Which table gives a group isomorphic to the group $U_{4}$, so that we know the binary operation defined by the table is associative?

(c) Show that the group given by one of the other tables is structurally the same as the group in Exercise 14 of the book for one particular value of $n$, so that we know that the operation defined by that table is associative also.

Solution
Table 1:

(1)
\begin{array} {c||c|c|c|c} * & e & a & b & c \\ \hline e & e & a & b & c \\ \hline a & a & e & c & b \\ \hline b & b & c & e & a \\ \hline c & c & b & a & e \\ \end{array}

Table 2:

(2)
\begin{array} {c||c|c|c|c} * & e & a & b & c \\ \hline e & e & a & b & c \\ \hline a & a & e & c & b \\ \hline b & b & c & a & e \\ \hline c & c & b & e & a \\ \end{array}

Table 3:

(3)
\begin{array} {c||c|c|c|c} * & e & a & b & c \\ \hline e & e & a & b & c \\ \hline a & a & b & c & e \\ \hline b & b & c & e & a \\ \hline c & c & e & a & b \\ \end{array}

Table 1 is different from 2 and 3 in that each element is its own inverse.

(a)

The symmetry in the main diagonal of each table shows that all groups of order $4$ are commutative.

(b)

Table 3 gives $U_{4}$ if $e = 1$, $a = i$, $b = -1$, and $c = -i$.

(c)

Let $n = 2$. Now, there are four $2 \times 2$ diagonal matrices with entries $\pm1$, which are

(4)
\begin{align} E = \begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}, A = \begin{bmatrix} -1 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}, B = \begin{bmatrix} 1 & 0 \\[0.3em] 0 & -1 \\[0.3em] \end{bmatrix}, C = \begin{bmatrix} -1 & 0 \\[0.3em] 0 & -1 \\[0.3em] \end{bmatrix} \end{align}

These matrices can be used to make up Table 1.