Find the "addition table" and "multiplication table" for a $6$-hour clock arithmetic.

$\displaystyle{\begin{array}{c|cccccc} + & 0 & 1 & 2 & 3 & 4 & 5 \\\hline 0 & 0 & 1 & 2 & 3 & 4 & 5 \\ 1 & 1 & 2 & 3 & 4 & 5 & 0 \\ 2 & 2 & 3 & 4 & 5 & 0 & 1 \\ 3 & 3 & 4 & 5 & 0 & 2 & 2 \\ 4 & 4 & 5 & 0 & 1 & 3 & 3 \\ 5 & 5 & 0 & 1 & 2 & 4 & 4 \end{array}}$ $\displaystyle{\begin{array}{c|cccccc} \times & 0 & 1 & 2 & 3 & 4 & 5 \\\hline 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 2 & 3 & 4 & 5 \\ 2 & 0 & 2 & 4 & 0 & 2 & 4 \\ 3 & 0 & 3 & 0 & 3 & 0 & 3 \\ 4 & 0 & 4 & 2 & 0 & 4 & 2 \\ 5 & 0 & 5 & 4 & 3 & 2 & 1 \end{array}}$

Find the multiplicative inverse of $7$ in a $11$-hour clock arithmetic.

There are more elegant ways (using the branch of mathematics known as number theory) to find such multiplicative inverses, especially when the "clock" has a large number of hours on it -- however, $11$ is a fairly small number of hours, so we can find the multiplicative inverse quickly, even using the following "brute-force" method:

Computing products $1 \cdot 7, 2 \cdot 7, 3 \cdot 7$, etc, all in this $11$-hour clock arithmetic, we stop upon seeing the first result equal ot $1$:$\def\ss{\kern-15pt}$ $$\begin{array}{ccccccccc} n \cdot 7\kern-7pt &=&\kern-7pt 7, \quad 2 \cdot 7\kern-7pt &=&\kern-7pt 3, \quad 3 \cdot 7 \kern-7pt&=&\kern-7pt 10, \quad 4 \cdot 7 \kern-7pt&=&\kern-7pt 6,\\ 5 \cdot 7\kern-7pt &=&\kern-7pt 2, \quad 6 \cdot 7\kern-7pt &=&\kern-7pt 9, \quad 7 \cdot 7 \kern-7pt&=&\kern-7pt 5, \quad 8 \cdot 7 \kern-7pt&=&\kern-7pt 1 \end{array}$$ Thus, $7^{-1} = 8$ in an $11$-hour clock arithmetic.

Determine which elements don't have multiplicative inverses in a $6$-hour clock arithmetic.

After having fleshed out a "multiplication table" for $6$-hour arithmetic (see below), this question is quickly answered. Given that $x$ and $x^{-1}$ are inverses precisely when $x \cdot x^{-1} = 1 = x^{-1} \cdot x$, we simply look for rows (and columns) that don't contain $1$ (the multiplicative identity). In this way, we see that $0$, $2$, $3$, and $4$ don't have multiplicative inverses.

$$\begin{array}{c|cccccc} \times & 0 & 1 & 2 & 3 & 4 & 5 \\\hline 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 2 & 3 & 4 & 5 \\ 2 & 0 & 2 & 4 & 0 & 2 & 4 \\ 3 & 0 & 3 & 0 & 3 & 0 & 3 \\ 4 & 0 & 4 & 2 & 0 & 4 & 2 \\ 5 & 0 & 5 & 4 & 3 & 2 & 1 \end{array}$$Below is the multiplication table for a $10$-hour clock arithmetic. Does the smaller set $S = \{1,3,7,9\}$ form a group using the same operation? If it does, is it an abelian group? Explain the reasoning behind both of your answers.

$$\begin{array}{c|cccccccccc} \times & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\\hline 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 2 & 0 & 2 & 4 & 6 & 8 & 0 & 2 & 4 & 6 & 8 \\ 3 & 0 & 3 & 6 & 9 & 2 & 5 & 8 & 1 & 4 & 7 \\ 4 & 0 & 4 & 8 & 2 & 6 & 0 & 4 & 8 & 2 & 6 \\ 5 & 0 & 5 & 0 & 5 & 0 & 5 & 0 & 5 & 0 & 5 \\ 6 & 0 & 6 & 2 & 8 & 4 & 0 & 6 & 2 & 8 & 4 \\ 7 & 0 & 7 & 4 & 1 & 8 & 5 & 2 & 9 & 6 & 3 \\ 8 & 0 & 8 & 6 & 4 & 2 & 0 & 8 & 6 & 4 & 2 \\ 9 & 0 & 9 & 8 & 7 & 6 & 5 & 4 & 3 & 2 & 1 \\ \end{array}$$Yes! It does form a group. To see this, consider its "multiplication" table below (which can be quickly obtained from the one above, by erasing the rows and columns corresponding to numbers

*not*in set $S = \{1,3,7,9\}$. $$\begin{array}{c|cccc} \times & 1 & 3 & 7 & 9 \\\hline 1 & 1 & 3 & 7 & 9 \\ 3 & 3 & 9 & 1 & 7 \\ 7 & 7 & 1 & 9 & 3 \\ 9 & 9 & 7 & 3 & 1 \end{array}$$ Then, note:$S$ is

*closed*with respect to our $10$ hour "multiplication", as we only see $1$s, $3$s, $7$s, and $9$s in the body of the smaller table.*Associativity*held in $10$-hour clock arithmetic, so given the table above involves the exact same calculations and is closed, associativity must also hold here.The

*identity*continues to be $1$ given that both the first row and column of products match the numbers being multiplied by $1$ to produce them.- Finally,
*inverses*for all elements exist as there is a $1$ in each row and column.

Further, we know this is an

*abelian*group given that there is a symmetry of products across the (falling) diagonal in the table. That is to say, for every $a \cdot b$ product (where both $a$ and $b$ are in $S = \{1,3,7,9\}$) equals its corresponding $b \cdot a$ product.Below is the addition table for a $10$-hour clock arithmetic. Does the smaller set $T = \{0,2,4,6,8\}$ form a group using the same operation? If it does, is it an abelian group? Explain the reasoning behind both of your answers. $$\displaystyle{\begin{array}{c|cccccccccc} + & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\\hline 0 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 1 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 0 \\ 2 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 0 & 1 \\ 3 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 0 & 1 & 2 \\ 4 & 4 & 5 & 6 & 7 & 8 & 9 & 0 & 1 & 2 & 3 \\ 5 & 5 & 6 & 7 & 8 & 9 & 0 & 1 & 2 & 3 & 4 \\ 6 & 6 & 7 & 8 & 9 & 0 & 1 & 2 & 3 & 4 & 5 \\ 7 & 7 & 8 & 9 & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ 8 & 8 & 9 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 9 & 9 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \end{array}}$$

First, let us find the addition table for $T = \{0,2,4,6,8\}$, by erasing the un-needed rows and columns of the table above: $$\begin{array}{c|ccccc} & 0 & 2 & 4 & 6 & 8 \\\hline 0 & 0 & 2 & 4 & 6 & 8 \\ 2 & 2 & 4 & 6 & 8 & 0 \\ 4 & 4 & 6 & 8 & 0 & 2 \\ 6 & 6 & 8 & 0 & 2 & 4 \\ 8 & 8 & 0 & 2 & 4 & 6 \end{array}$$ Then, note:

$T$ is

*closed*with respect to $10$-hour "addition", as we only see elements of $T$ in the body of the table.*Associativity*holds as it held in the larger table.The

*identity*is $0$, as adding this to any value (even on a clock) leaves it unchanged.Every element has an

*inverse*as we see the identity $0$ in every row and column.

Thus, $T$ with addition on a $10$-hour clock forms a group.

Further, given the symmetry seen along the (falling) diagonal where every $ab = ba$ for elements $a$ and $b$ in $T$, this is an

*abelian*group.Does the set $T = \{0,2,4,6,8\}$ form a commutative ring with respect to addition and multiplication on a $10$-hour clock?

No. Notice that multiplication, if restricted to the elements of $T$, has no multiplicative identity.