Prove that $G$ be a group under $\oplus$ that defined by $\bar{a} \oplus \bar{b} = \bar{a} \times_7 \bar{b} +_7 \bar{a} +_7 \bar{b}$.

Question

In this problem, I stuck when find the inverse of each element in $G$. Please help me at least give a hint so that I can solve this problem clearly.

Problem

Let $\mathbb{Z}_7$ be a group under $+_7$ and $\mathbb{Z}_{7}^{*}$ be a group under $\times_7$ where $\mathbb{Z}_{7}^{*} = \lbrace \bar{a} \in \mathbb{Z}_7 \mid \bar{a} \neq \bar{0} \rbrace$. Let a nonempty set $G$ that defined as $G = \lbrace \bar{a} \in \mathbb{Z}_7 \mid \bar{a} \neq \bar{6} \rbrace$ and a binary operation $\oplus$ on $G$ that defined by $$\bar{a} \oplus \bar{b} = \bar{a} \times_7 \bar{b} +_7 \bar{a} +_7 \bar{b}$$ for all $\bar{a},\bar{b} \in \mathbb{Z}_7$. Prove that $G$ is a group under $\oplus$.

My Solution.

It's easy to show that $\oplus$ is an associative binary operation.

Now, $\bar{0} \in G$. Then, $\bar{0} \oplus \bar{a} = \bar{0} \times_7 \bar{a} +_7 \bar{0} +_7 \bar{a} = \overline{0+a} = \bar{a} = \bar{a} \oplus\bar{0}$. Thus, $\bar{0}$ be an identity element of $G$.

Next, we'll find the inverse. Let $\bar{a}, \bar{m} \in G$ where $\bar{m}$ be the inverse of $\bar{a}$. Then, $\bar{0} = \bar{m} \oplus \bar{a} \Rightarrow \bar{m} = -\frac{\bar{a}}{\bar{a}+1} \notin G$.

I get stuck. Please help at least give me some hint. Thanks!

Answer 1

Associativity follows from that of multiplication & addition of integers.

It is clear that, indeed, $\bar{0}$ is the identity.

The Cayley table is then computed, with the help of $\color{blue}{\text{commutativity}}$ of multiplication & addition of integers (and hence of $\oplus$), as follows:

$$\begin{array}{c|cccccc} \oplus & \bar{0} & \bar{1} & \bar{2} & \bar{3} & \bar{4} & \bar{5} \\ \hline \bar{0} & \bar{0} & \bar{1} & \bar{2} & \bar{3} & \bar{4} & \bar{5} \\ \bar{1} & \color{blue}{\bar{1}} & \bar{3} & \bar{5} & \bar{0} & \bar{2} & \bar{4} \\ \bar{2} & \color{blue}{\bar{2}} & \color{blue}{\bar{5}} & \bar{1} & \bar{4} & \bar{0} & \bar{3} \\ \bar{3} & \color{blue}{\bar{3}} & \color{blue}{\bar{0}} & \color{blue}{\bar{4}} & \bar{1} & \bar{5} & \bar{2} \\ \bar{4} & \color{blue}{\bar{4}} & \color{blue}{\bar{2}} & \color{blue}{\bar{0}} & \color{blue}{\bar{5}} & \bar{3} & \bar{1} \\ \bar{5} & \color{blue}{\bar{5}} & \color{blue}{\bar{4}} & \color{blue}{\bar{3}} & \color{blue}{\bar{2}} & \color{blue}{\bar{1}} & \bar{0} \end{array},$$

from which one can deduce that

$$\begin{align} \bar{1}^{-1}&=\bar{3},\\ \bar{2}^{-1}&=\bar{4},\\ \bar{3}^{-1}&=\bar{1},\\ \bar{4}^{-1}&=\bar{2},\,\text{ and}\\ \bar{5}^{-1}&=\bar{5}. \end{align}$$

Closure is also implied by the table.

Thus $(G, \oplus)$ is a group.

Answer 2

You are confusing the inverse in $\mathbb R$ and the inverse in $\mathbb Z_7^*$. To compare:

\begin{array}{c | c | c} x & x^{-1}\, \text{in}\,\mathbb R & x^{-1}\, \text{in}\,\mathbb Z_7^* \\ \hline 1 & 1 & 1\\ 2 & 1/2 & 4 \\ 3 & 1/3 & 5 \\ 4 & 1/4 & 2 \\ 5 & 1/5 & 3 \\ 6 & 1/6 & 6 \end{array}

To be fair, elements of $\mathbb R$ and elements of $\mathbb Z_7^*$ are fundamentally different when it comes to algebra, so using the same symbols for them tends to mislead beginners. That's why on the introductory level, we use something like $\bar n$ to denote the equivalence class of $n\in\mathbb Z$ under some relation, here: $n\sim m \iff n-m\in 7\mathbb Z.$

Let me rewrite the above table using this notation:

\begin{align} \bar 1 \times_7 \bar 1 = \bar 1 &\implies (\bar 1)^{-1} = \bar 1,\\ \bar 2 \times_7 \bar 4 = \bar 8 = \bar 1 &\implies (\bar 2)^{-1} = \bar 4, (\bar 4)^{-1} = \bar 2,\\ \bar 3 \times_7 \bar 5 = \overline {15} = \bar 1 &\implies (\bar 3)^{-1} = \bar 5, (\bar 5)^{-1} = \bar 3,\\ \bar 6 \times_7 \bar 6 = \overline {36} = \bar 1 &\implies (\bar 6)^{-1} = \bar 6.\\ \end{align}

I could have just guessed those, or better yet, I could have made the multiplication table for $\mathbb Z_7^*$:

\begin{array}{ c| c c c c c } \times_7 & \bar 1 & \bar 2 & \bar 3 & \bar 4 & \bar 5 & \bar 6\\ \hline \bar 1 & \bf{\color{red}{\bar 1}} & \bar 2 & \bar 3 & \bar 4 & \bar 5 & \bar 6\\ \bar 2 & \bar 2 & \bar 4 & \bar 6 & \bf{\color{red}{\bar 1}} & \bar 3 & \bar 5\\ \bar 3 & \bar 3 & \bar 6 & \bar 2 & \bar 5 & \bf{\color{red}{\bar 1}} & \bar 4\\ \bar 4 & \bar 4 & \bf{\color{red}{\bar 1}} & \bar 5 & \bar 2 & \bar 6 & \bar 3\\ \bar 5 & \bar 5 & \bar 3 & \bf{\color{red}{\bar 1}} & \bar 6 & \bar 4 & \bar 2\\ \bar 6 & \bar 6 & \bar 5 & \bar 4 & \bar 3 & \bar 2 & \bf{\color{red}{\bar 1}} \end{array}

As you can see, each row and column has precisely one $\bf{\color{red}{\bar 1}}$, so every element of $\mathbb Z_7^*$ has unique inverse.

To solve your problem, we have $G = \{\bar 0, \bar 1, \bar 2,\bar 3,\bar 4,\bar 5\}$. Now make a multiplication table like above, only use your operation $\oplus$ and look for $\bf{\color{blue}{\bar 0}}$.

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I strongly encourage you to do the above. The alternative approach is less elementary.

Notice two things, $\bar a\oplus \bar b +_7 \bar 1 = (\bar a +_7 \bar 1)\times_7 (\bar b +_7 \bar 1)$ and if we add $\bar 1$ to all the elements of $G$, we get all the elements of $\mathbb Z_7^*$, i.e. $f\colon G\to \mathbb Z_7^*$, $f(\bar x) = \bar x +_7 \bar 1$ is a bijection with the inverse $g\colon \mathbb Z_7^*\to G$, $g(\bar x) = \bar x -_7 \bar 1$.

These two things tell me that there is a strong relationship between $G$ and $\mathbb Z_7^*$. Furthermore

$$f(\bar a\oplus \bar b) = \bar a\oplus\bar b +_7 \bar 1 = (\bar a+_7\bar 1)\times_7(\bar b+_7\bar 1) = f(\bar a)\times_7 f(\bar b)$$ and similarly $$g(\bar a\times_7 \bar b) = g(\bar a)\oplus g(\bar b).$$

(Compare this to your previous question and my answer there.)

Now, I can easily prove that $G$ is a group (I will lose the $\bar\cdot$, $\times_7$ and $+_7$ notation, even though that is what is meant):

1. $(a\oplus b)\oplus c + 1 = (a\oplus b + 1)(c + 1) = (a+1)(b+1)(c+1) = \ldots =a\oplus(b\oplus c) + 1$ $\implies (a\oplus b)\oplus c = a\oplus(b\oplus c),$
2. $a\oplus 0 + 1 = (a+1)(0 + 1) = a + 1 = (0+1)(a+1) = 0\oplus a + 1$ $\implies a\oplus 0 = a = 0\oplus a,$

This is where it gets trickier:

\begin{align} 1\cdot 1 = 1 \implies (0+1)(0+1) = 1 &\implies 0\oplus 0 + 1 = 1 \implies 0\oplus 0 =0,\\ 2\cdot 4 = 1 \implies (1+1)(3+1) = 1 &\implies 1\oplus 3 + 1 = 1 \implies 1\oplus 3 = 0,\\ 3\cdot 5 = 1 \implies (2+1)(4+1) = 1 &\implies 2\oplus 4 + 1 = 1 \implies 2\oplus 4 = 0,\\ 6\cdot 6 = 1 \implies (5+1)(5+1) = 1 &\implies 5\oplus 5 + 1 = 1 \implies 5\oplus 5 = 0.\\ \end{align}

So, there you have your inverses.

Essentially, we have established that $f$ and $g$ are group isomorphisms of $G$ and $\mathbb Z_7^*$, so they map inverses to inverses.