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I am taking a class in group theory (we began with basic number theory). I am trying to show that there is a homomorphism from $\mathbb{Z}$ to $\mathbb{Z}_n$. By this, I mean from the integers to the integers mod n. I know this is a standard proof that has already been asked, but I do not understand them so I am trying to figure it out myself.

To show that a mapping $f$ is homomorphic, you need to show that $f(xy)=f(x)f(y)$, but for this problem it's $f(x+y)=f(x)+f(y)$. With the division algorithm I can write $(x+y)=qn+r$ where $0\le r< n$ to get $f(x+y)=r$. For $x$ and $y$,we get $x=q_1n+r_1$, $y=q_2n+r_2$, so $f(x)=r_1$ and $f(y)=r_2$. This is where I am stuck, I do not understand how to show that $r=r_1+r_2$. Intuitively, I think that I am missing a basic modular arithmetic lemma, but I don't know what it is. (The proofs that I saw do not explicitly say what said lemma is.)

Musical Taco
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1 Answers1

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Writing $x = an + r_1$ and $y = bn + r_2$, we have, under addition modulo $n$, that:

$$x + y = an + bn + r_1 + r_2 = (a+b)n + (r_1 + r_2) \equiv r_1 + r_2 \text{ mod }n$$

Assuming that you are defining $f(x) := x \text{ mod }n$, the corresponding claim now follows from the fact that, in modular arithmetic, it is true that:

$$(r_1 + r_2) \text{ mod }n = (r_1 \text{ mod }n) + (r_2 \text{ mod }n)$$

In the equation above, the left hand side is $f(x + y)$ and the right hand side is $f(x) + f(y)$.

Benjamin Dickman
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  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. It's best for site health to delete this answer (which also minimizes community time wasted on dupe processing.) [once the question is clarified it will surely be a dupe, and your answer already is]. – Bill Dubuque Mar 17 '24 at 21:15
  • @BillDubuque I thought that by 1 )tagging this under solution verification and 2) providing exactly where I am confused with the proof I proved, it wouldn't be consider a dupe. – Musical Taco Mar 17 '24 at 21:19
  • @Musical If the above is the sort of answer you seek then it is definitely a duplicate (of tens if not hundreds of prior posts). Adding an SV tag does not stop it from being a dupe - see the SV tag info. See here in the first dupe for a more conceptual method. – Bill Dubuque Mar 17 '24 at 21:30
  • @BillDubuque I did check the SV tag. The two posts that you recommended did not specifically help me when trying to figure out my proof. I have no idea what $((a\bmod n)+(b\bmod n))\bmod n)$ means. I hit a wall with my last step. I made a guess that I needed $a\bmod n+b\bmod n=(a+b)\bmod n$, which I did not know was true, and it just so happened that that was what I needed for this specific proof. Nowhere in my textbook does it mention that lemma so I did not know if it was true. – Musical Taco Mar 17 '24 at 21:37
  • @Musical What precisely does $,a\bmod n,$ denote in your context? Is it an integer, or a coset or congruence class, i.e. what is your definition of $,\Bbb Z_n?,$ As I commented on the question, you should specify this crucial context. – Bill Dubuque Mar 17 '24 at 21:43
  • @Musical Beware that the 2nd displayed formula in the answer is incorrect if you interpret as pure modular arithmetic, e.g. $,0 = (1+2)\bmod 3 \neq (1\bmod 3)+ (2\bmod 3) = 3.,$ That's why we need an extra final mod on the RHS, then we correctly get $,0 = 3\bmod 3 =0.,$ Or instead we could interpret the RHS addition as an operation in $,\Bbb Z_n,$ (vs. $\Bbb Z),,$ e.g. as coset addition. See also here on $!\bmod!$ the operation vs. congruence relation. – Bill Dubuque Mar 17 '24 at 22:06