Prove the associativity of $(\Bbb Z_7,\oplus)$.

Question

I am reading this Problem 14.4.1.

We want to prove that $(Z_7,\oplus)$ is group. I have difficulty proving associativity axiom. The solution reads

Associativity: Let $a\in\mathbb Z_7,$ $b\in\mathbb Z_7$ and $c\in\mathbb Z_7$. By Theorem 3.4.10 we only need to show $$(a+(b+c))\bmod 7 = ((a+b)+c)\bmod 7.$$ This holds since $a+(b+c)=(a+b)+c$ for all integers $a$, $b$, and $c$ by the associative property of the integers. Hence $\oplus$ is associative.

Theorem $3.4.10$ states that

$$\boxed{ \textbf{Therorem 3.4.10.} \text{ Let $a$ and $b$ be integers, and let $m$ be a natural number. Then}\\ (a+b)\bmod m=\big((a\bmod m)+(b\bmod m)\big)\bmod m }$$

I am having hard time understanding why from above theorem follows that we only need to show that $(a+(b+c))\pmod{7} =((a+b)+c)\pmod{7}$?

Can you explain this part?

We need to show that $a\oplus (b \oplus c))$ = $(a \oplus b) \oplus c$

$a\oplus (b \oplus c))=a\oplus((b+c)\bmod 7) =(a+((b+c)\bmod7))\bmod7$

As Shaun said, don't use pictures, since they might not be accessible for everyone and cannot be searched. I edited your question and replaced the pictures by their texts. Please check that I didn't change anything unintentionally. — anankElpis, Feb 09 '22 at 16:17
As far as I could tell, @BillDubuque, that wasn't a duplicate. — Shaun, Feb 09 '22 at 18:10
@Shaun Please read the linked Key Idea. Your answer duplicates that answer. Please reclose it. — Bill Dubuque, Feb 09 '22 at 18:11
The question here is about associativity, @BillDubuque, not index laws. — Shaun, Feb 09 '22 at 18:13
@SHain The Key Idea explains why it is valid to "ersase" (or insert) mod operations in any "polynomial" expression (i.e. any composition of sums and product operations), which includes the polynomials $(a+b)+c = a+(b+c)$ — Bill Dubuque, Feb 09 '22 at 18:17

score 1 · Accepted Answer · answered Feb 09 '22 at 16:33

The point of the theorem 3.4.10 is that you can take the $\bmod$ additionally at any point in your calculation without changing the result. This should make it clear intuitively that the statement is true.

If you want to be very formal, you can continue your calculation \begin{align}a\oplus(b\oplus c)&=\big(a+(b+c)\bmod 7\big)\bmod 7\\ &\overset{(1)}=\Big(a\bmod 7+\big((b+c)\bmod7\big)\bmod 7\Big)\bmod7\\ &\overset{(2)}=\big(a\bmod7+(b+c)\bmod7\big)\bmod7\\ &\overset{(3)}=(a+(b+c))\bmod7, \end{align} which is exactly the claim the solution makes (after you do the same exact calculation for $(a\oplus b)\oplus c$). For (1) and (3), we used the theorem, for (2) notice that $\bmod 7$ is applied twice to $b+c$ in the second line, which is unnecessary.

score -1 · Answer 2 · answered Feb 09 '22 at 16:08

If what you are looking for is an intuitive approach, you can appreciate if you imagined a clock with only 7 numbers; let those numbers be $0,\ldots ,6$. Adding 2 different hours $h_1,h_2$, no matter in which order, will always give you the same hour in the clock. This is what theorem 3.4.10 states.

Now, from a formal definition, associativity of an operation $*$ in a group is defined as $a*(b*c)=(a*b)*c$, and that is what "suffices" you to prove. However, if what you mean is that why does it also imply $a+(b+c)\mod 7 = a+b+c\mod 7$ comes from extending your theorem 3.4.10 to that particular case.

Hope it helps.

Prove the associativity of $(\Bbb Z_7,\oplus)$.

2 Answers2