# Showing the group of integers modulo m with multiplication is a group

I've just started group theory and I don't know how to show this. Are you supposed to take examples of two elements from the group and an example modulo m (say 2 and 5, and modulo 3) and show the group axioms hold? That doesn't seem sufficient to me, it seems like I should be taking $a, b, c$ as elements of the group and proving it in general. I can see how to show that G has an identity element in that case but how do I show it has an inverse and that associativity holds?

Let $a, b, c \in G, G=(\mathbb{Z}/m\mathbb{Z})$

Identity:

$a * 1 = a$ so $1$ is the identity

Inverse:

I don't know how to show this...

Associativity:

I don't know how to show this either...

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The integers modulo $m$ under multiplication is not a group. You either want the integers modulo $m$ under addition or the integers relatively prime to $m$ under multiplication modulo $m$. –  JSchlather Oct 2 '12 at 13:34

## 3 Answers

The problem is not well stated, as you cannot have $0$ in your group. It should ask that the integers in $\mathbb Z/m\mathbb Z$ that are coprime to $m$ be a group under multiplication. Associativity comes because it is true in $\mathbb Z$. To find the inverse you need Bézout's lemma

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Why can I not have $0$ in my group? –  dukenukem Oct 2 '12 at 17:15
@dukenukem: because it doesn't have an inverse. If $m$ is $15$, neither will $3, 5, 6, 9, 10, 12$ which is why you want the numbers coprime to $m$. –  Ross Millikan Oct 2 '12 at 17:48

$\mathbb Z/m\mathbb Z$ is a ring in general: it consists of an additive group (it is indeed a group, with respect to $+$: $\ 0$ is the unit element and $-x$ is the inverse of $x$, now $-x\equiv m-x$), and a multiplicative monoid (with respect to $\cdot)$ has a unit, and the product is associative). Now multiplication is also commutative but it's not necessary for being a ring. And the distirbutive law connects these two.

$0$ never will have a multiplicative inverse. Exactly those $a$ will have such, which are relatively prime to $m$. So, if $m$ is prime, then $\mathbb Z/m\mathbb Z$ is a field, i.e. every nonzero element has a multiplicative inverse.

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For every a s.t. gcd(a,n) = 1 there exists a multiplicative inverse; use Extended Euclidean Algorithm to find it.

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