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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?

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$a^{-1}=1-a=1+6a$ that is $a^{-1}=(1+6a)$ mod 7 isnt it ? –  pritam Jul 8 '12 at 9:11
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No, it's not acceptable to state that $a+1$ is the inverse of $a$, since it's not true. For example, take $a=1$. Then $a\odot (a+1)=1\odot 2=1+2+3=6\neq 4$. –  Chris Eagle Jul 8 '12 at 9:11
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Note you did not prove the alleged group has an identity. What you did prove is that if it does have an identity, then the identity is 4. It's unclear if you have the same conceptual mistake for your proofs of the other properties. –  Hurkyl Jul 8 '12 at 10:33
    
@Hurkyl what about the associativity, commutativity and identity element? If my calculation is right and all of those apply along with the presence of the inverse elements isn't that enough to say that $(\mathbb Z_7, \odot)$ is an abelian group? If not what am I missing? –  haunted85 Jul 8 '12 at 10:40
    
@haunted85: Your proof outline is fine: the individual steps are true (once you've corrected your typo for inverses) and the conclusion holds. However, the way you did the "prove it has an identity" step is backwards, and I suspect you did the other two steps backwards as well (but I can't know for sure without knowing your actual thoughts about what you were doing). –  Hurkyl Jul 8 '12 at 10:46
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3 Answers 3

up vote 1 down vote accepted

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

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You have $\rm\:a + a^{-1} = 1,\:$ so $\rm\: a^{-1} = 1 - a.\:$ Note that all of your equations should be connected by arrows going both ways, i.e. $\iff\!\!,\:$ since you need to prove both 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 the operands by applying $\,\ell^{-1}n\, =\, n\!+\!3,\,$ then perform the normal operation, then label the result, i.e.

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

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

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

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

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

Note that the addition table on the right is the table for the operation of addition mod $7$, except the rows and colums have been reordered (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 after a (renaming) bijection is applied to the elements. The notion of isomorphism is defined so that the algebraic structure is determined completely by the 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.

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).

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

enter image description here

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pictures, graphs, tables, etc always help! + –  amWhy Mar 9 '13 at 0:09
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