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I received some help and direction on this from some users a few days ago, and have tried to take that information and craft it into something proofy. I would appreciate general suggestions, edits, and verification for for my proof-sketch.

Please note-- I am new to proof writing, and preparing for Real Analysis in the Fall. This proof required that I read a few sections I have not been lectured on (integers modulo $n$, binary operations, and groups). I have been taking an into to abstract math course over the summer.

Here is my "proof sketch" for this problem. I am still stuck on the last part (see below).

Proof (direct): We will show that $(\mathbb{Z}_n, +)$ is a group by showing that it follows the Associative Law, Existence of Identity, and Existence of Identity properties that form a group.

Associative Law: Let [$a$], [$b$], [$c$] $\in \mathbb{Z}_n$. By the Associative law of Set Addition, we know that $\\ [a]+[b] = [a + b]$. Observe:

$[a]+([b]+[c])=[a]+[b+c]=[a+(b+c)]=[(a+b)+c]=([a]+[b])+[c]$, Thus the equivalence classes of $\mathbb{Z}_n$ hold under Set Addition, and follow associative laws.

Existence of an Identity: Assume that $[0]=\left\{ x \in \mathbb{Z} \mid x \equiv 0(\text{mod} \ n) \right\}$ and let $[a] \in \mathbb{Z}_n$ such that $[a]+[0]=[0]+[a]=[a]$. This indicates that $(\mathbb{Z}_n,+)$ has the existence of an identity.

Existence of Inverse: Let $[a] \in \mathbb{Z}_n$ such that $[a]\equiv k(\text{mod} \ n)$, where $k \in \mathbb{Z}$.

Okay, I don't know how to show this last part. I have been playing with rearranging modular information, but am not certain if that is how I want to go about setting it up. I've been looking into information such as $[a]\equiv k (\text{mod}n)$ and $[b]\equiv j (\text{mod}n)$, and thought I could set out to rearrange this to show the inverse (since $k^-1=n-k$), but this isn't getting me anywhere. I realize I need to ultimately show that the inverse of $[a]$ is $[-a]$ and that $[-a] \in \mathbb{Z}_n$... but I can't quite get there.

PS- I have posted a previous question, in which I ask for help with this. I am not sure of proper MS etiquette when it comes to updating questions/information to ask for help, so let me know if I should approach this in a different way.

Mr. Meeseeks
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  • $-a \equiv n-a\pmod n$. Try proving that $[a]+[n-a] = [0]$. – Cameron Williams Jul 26 '15 at 16:39
  • I'm not following some of this. For the associative law, don't you want to show a+(b+c) = (a+b) + c, and you can do this by showing modulo arithmetic is associative. For existence of identify, show that 0+a = a = a+0. Existence of inverse, pick a and then show -a is its inverse. – user247608 Jul 26 '15 at 18:25
  • @cockneywanker I was first trying to show associative by using modulo arithmetic, but was then told that was not necessary. I couldn't get it to work with modulo arithmetic, so if you have a suggestion... – Mr. Meeseeks Jul 26 '15 at 18:28
  • @cockneywanker Similarly, I realize that I need to choose $[a]$ and show that [-a] is it's inverse, but I don't know under what property to start to solve this (do I need to use modulo arithmetic?) – Mr. Meeseeks Jul 26 '15 at 18:29
  • You have to show the associativity law works on the equivalence class [a], [b], [c]. Pick an element from [a]+([b]+[c]), this is of the form x = a+(b+c) which by the associativity law x = (a+b)+c so x is also a member of ([a]+([b]+[c]), similarly you can do the reverse so that shows [a]+([b]+[c]) = ([a]+([b]+[c]). – user247608 Jul 26 '15 at 19:23
  • for the inverse, you can do a similar thing. pick an element from [a] and an element from [-a], call them x and y, so x = a % n and y = (-a) % n. Look at x+y , = (a%n) + (-a%n) = (a + -a) % n = 0 % n = 0, so [a]+[-a] is the equivalence class of the identity element. You also have to do the reverse [-a]+[a], so that shows [-a] is the inverse of [a] - does that help? – user247608 Jul 26 '15 at 19:27
  • I should probably be more accurate and say x = a%n + kn and y = -a%n + ln , a representative of [a+-a] would contain elements such as z = (a%n) + (-a%n) + kn + ln which can be simplified to z = (a+-a)%n + (k+l)n and further simplified to z = 0 + mn so z is a member of [0] hence [a]+[-a] = [0], etc. – user247608 Jul 26 '15 at 19:38
  • @cockneywanker I haven't seen the " % " notation yet, is this a symbol for "divides"? – Mr. Meeseeks Jul 26 '15 at 19:44
  • Sorry % is the modulus operator used in computing languages, A % B is the remainder of A divided by B. – user247608 Jul 26 '15 at 20:34

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