Check if $(\mathbb Z_7, \odot)$ is an abelian group, issue in finding inverse element

Question

Take $\mathbb Z_7$ and the operation $\odot$ defined on it as follows $\forall a,b \in \mathbb Z_7$:

$$\begin{aligned} a \odot b=a+b+3\end{aligned}$$

Check if $(\mathbb Z_7, \odot)$ is a group and in particular if it is an abelian group.

Associativity

It can be easily proved that $\odot$ is associative:

$$\begin{aligned} (a \odot b) \odot c = a \odot (b \odot c)\end{aligned}$$ $$\begin{aligned}(a + b + 3) \odot c = a \odot (b+c+3)\end{aligned}$$ $$\begin{aligned} (a + b + 3) + c +3 = a + (b+c+3)+3 \end{aligned}$$ $$\begin{aligned} a + b + c +6 = a +b+c+6 \end{aligned}$$

Commutativity

$\odot$ is commutative:

$$\begin{aligned} a \odot b = b \odot a \end{aligned}$$ $$\begin{aligned} a + b + 3 = b +a+3 \end{aligned}$$

Identity element

There is an identity element for $(\odot, \mathbb Z_7)$

$$\begin{aligned} a \odot \mathbb 1_{\mathbb Z_7} = a \end{aligned}$$ $$\begin{aligned} a + e + 3 = a \end{aligned}$$ $$\begin{aligned} e + 3 = 0 \end{aligned}$$ $$\begin{aligned} e = 4 \end{aligned}$$

and similarly if we repeat the calculation for $\mathbb 1_{\mathbb Z_7} \odot a$, still $e=4$

What I am struggling with is the inverse element, how to check if there is one. I know that it should be:

$$\begin{aligned} a \odot a^{-1} = e \\ a + a^{-1} + 3 = e \\ a + a^{-1} + 3 = 4\end{aligned}$$

Is it acceptable to state $a^{-1} = a+1$ (and similarly for $a^{-1} \odot a = e$)? If so, have I succeeded in proving $(\odot, \mathbb Z_{7})$ to be an abelian group?

Answer 1

We have $\rm\:a + a^{-1} = 1,\:$ so $\rm\: a^{-1} = 1 - a.\:$ All of your equations should be connected by arrows going both ways $(\!\!\iff\!\!)$ since you need to prove equivalences (necessity and sufficiency).

Here the group structure arises simply from renaming (or labeling) the elements of the additive group $\,\Bbb Z/7\,$ via the "label" bijection $\rm\:\ell\, n := n-3,\:$ i.e. by naming or labeling each natural mod $7\,$ by the natural congruent to $\rm\,n\!-\!3.\,$ To perform an operation on labels, we first unlabel arguments by applying $\,\mu\,n:=\ell^{-1}n\, =\, \rm n\!+\!3,\,$ then perform the operation in $\Bbb Z/7,\,$ then label the result, i.e.

$$ a \oplus b\ :=\ \ell\,(\ell^{-1}a\, +\, \ell^{-1}b)\ =\rm\, -3 + ((a\!+\!3) + (b\!+\!3))\ =\ a+b+3 $$

$$ \ominus\, a\ :=\ \ell(-\,\ell^{-1}a)\ =\rm\ {-}3+ (-(a\!+\!3))\ =\ {-}6-a\ =\ 1-a\qquad $$

Thus we have $\, \mu(a \oplus b)\, =\, \mu\,a + \mu\, b,\ $ and $\, \mu(\ominus a)\, =\, -\mu\, a\ $ so $\,\mu\,$ is a bijective group homomorphism, hence an isomorphism. In more technical language we say that we have transported the group structure along the bijection $\,\ell\,$ (or $\,\mu = \ell^{-1})$.

For example the equation $\,5+ 6 = 4\,$ transports to $\,\ell\, 5 \oplus \ell\, 6 = \ell\, 4,\,$ i.e. $\it\, 2 \oplus 3 = 1,\,$ and $\,-(5)\, =\, 2\,$ goes to $\,\ominus(\ell\,5)\, =\, \ell\, 2,\,$ i.e. $\it\,\ominus\:\!2 = 6.\,$ Transporting the whole addition table yields

$$\begin{array}{|c|c|c|c|c|c|c|c|} \hline \color{#C00}\oplus &\it\color{#C00}0 &\it\color{#C00}1 &\it\color{#C00}2 &\it\color{#C00}3 &\it\color{#C00}4 &\it\color{#C00}5 &\it\color{#C00}6 \\ \hline \it\color{#C00} 0 &\it 3 &\it 4 &\it 5 &\it 6 &\it 0 &\it 1 &\it 2 \\ \hline \it\color{#C00} 1 &\it 4 &\it 5 &\it 6 &\it 0 &\it 1\, &\it 2 &\it 3 \\ \hline \it\color{#C00} 2 &\it 5 &\it 6 &\it 0 &\it 1\, &\it 2 &\it 3 &\it 4 \\ \hline \it\color{#C00} 3 &\it 6 &\it 0 &\it 1\, &\it 2 &\it 3 &\it 4 &\it 5 \\ \hline \it\color{#C00} 4 &\it 0 &\it 1\, &\it 2 &\it 3 &\it 4 &\it 5 &\it 6 \\ \hline \it\color{#C00} 5 &\it 1\, &\it 2 &\it 3 &\it 4 &\it 5 &\it 6 &\it 0 \\ \hline \it\color{#C00} 6 &\it 2 &\it 3 &\it 4 &\it 5 &\it 6 &\it 0 &\it 1\, \\ \hline \end{array} \ \ \begin{array}{c} \xrightarrow[\large \ \it N\ \to\,\rm N+3\ ]{\large \rm unlabel\,\ \mu} \\ \\ \\ \xleftarrow[\large \ \it N-3\ \leftarrow\, \rm N\ ]{\large \rm label\,\ \ell} \end{array}\ \ \begin{array}{|c|c|c|c|c|c|c|c|} \hline \color{#C00}+ &\color{#C00} 3 &\color{#C00} 4 &\color{#C00} 5 &\color{#C00} 6 &\color{#C00} 0 &\color{#C00} 1 &\color{#C00} 2 \\ \hline \color{#C00}3 & 6 & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline \color{#C00}4 & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \color{#C00}5 & 1 & 2 & 3 & 4 & 5 & 6 & 0 \\ \hline \color{#C00}6 & 2 & 3 & 4 & 5 & 6 & 0 & 1 \\ \hline \color{#C00}0 & 3 & 4 & 5 & 6 & 0 & 1 & 2 \\ \hline \color{#C00}1 & 4 & 5 & 6 & 0 & 1 & 2 & 3 \\ \hline \color{#C00}2 & 5 & 6 & 0 & 1 & 2 & 3 & 4 \\ \hline \end{array}$$

Note that the addition table on the right is that for addition mod $7$, except the rows and columns have been permuted (shifted by $3$). Thus the two addition tables are essentially the same, i.e. they differ only in the names chosen for the elements. This is the sense of isomorphism that is captured by the notion of isomorphic groups, i.e. the two groups have exactly the same operation tables (up to order), i.e. they are identical after a (renaming) bijection is applied to the elements. This notion of algebraic isomorphism (sameness) is defined so that an algebraic structure is determined completely by its operation tables, i.e. the only properties of the elements that we care about algebraically are how the elements relate to each other under the operations. Any other (internal) structure the elements may possess (names, set-theoretic representation, etc), play no role algebraically (this is elaborated here).

Similarly we can transport the group structure along any permutation $\,\ell\,$ of $\,\Bbb Z/7$, and we can label or index any finite group by natural numbers (e.g. which might be addresses in computer memory, where (un)label operations amounts to memory (de)references).

Answer 2

I just added a table of your group. Maybe it helps you a little bit. Other comments and answer is enough for your question.

Answer 3

The inverse is $a^{-1}=1-a$, as $a\odot (1-a)=a+(1-a)+3=4=e$.