1. ## modulo arithmetic proof

Afternoon guys, I'm stuck on this proof and I'd really appreciate it if someone could help solve it. ($\displaystyle Z$ = set of integers)

Question/Theorem

Let $\displaystyle Z^*_p$ be the set of nonzero integers modulo p. ($\displaystyle p \geq 2$) Prove $\displaystyle Z^*_p$ is a group with multiplication mod $\displaystyle p$ if and only if $\displaystyle p$ is a prime number.

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My (lame) effort so far:

We need to show:
(1) $\displaystyle Z^*_p$ is a group with multiplication mod $\displaystyle p$ $\displaystyle \Rightarrow$ $\displaystyle p$ is prime.

(2) if $\displaystyle p$ is prime, $\displaystyle Z^*_p$ is a group with multiplication mod $\displaystyle p$
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(1) $\displaystyle \forall x \in Z^*_p$, $\displaystyle \exists x^{-1} \in Z^*_p$ such that $\displaystyle xx^{-1} = 1$

so $\displaystyle xx^{-1} = 1 \mod p$
$\displaystyle xx^{-1} -1 = 0 \mod p$
$\displaystyle \Rightarrow$ p divides $\displaystyle xx^{-1} -1$
$\displaystyle \Rightarrow \exists a \in Z$ such that $\displaystyle xx^{-1} - pa =1$

I don't know where to go from here, or indeed if this is the right way of approaching this.
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(2) if $\displaystyle p$ is prime, then any integer $\displaystyle q$ such that $\displaystyle 1 \leq q \leq p-1$ is coprime to $\displaystyle p$ and so $\displaystyle \exists$ n,m $\displaystyle \in Z$ such that $\displaystyle np + mq =1$

Now using that, we need to show the properties of a group are satisfied.

Showing associativity and identity (=1) are easy enough. But I don't know how to show an inverse exists. i.e. show that $\displaystyle \exists x^{-1} \in Z^*_p$ such that $\displaystyle xx^{-1} = 1$ using the above property of a prime number.

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Thank you.

2. Originally Posted by WWTL@WHL
Afternoon guys, I'm stuck on this proof and I'd really appreciate it if someone could help solve it. ($\displaystyle Z$ = set of integers).
Consider $\displaystyle \{ [0],[1],...,[p-1]\}$ we want to show if $\displaystyle [x]\not = [0]$ then there exists $\displaystyle [y]$ such that $\displaystyle [xy]=[1]$. This is equvalent to saying $\displaystyle xy\equiv 1(\bmod p)$. Since $\displaystyle p$ is prime it means $\displaystyle \gcd(x,p)=1$ since $\displaystyle [x]\not = [0]$. Thus, $\displaystyle ax+bp = 1$ for some integers $\displaystyle a,b$. This means that $\displaystyle ax\equiv 1(\bmod p)$ for some $\displaystyle a$. Thus, if we let $\displaystyle [y] = [a]$ then $\displaystyle [xy]=1$.

3. Originally Posted by ThePerfectHacker
Consider $\displaystyle \{ [0],[1],...,[p-1]\}$ we want to show if $\displaystyle [x]\not = [0]$ then there exists $\displaystyle [y]$ such that $\displaystyle [xy]=[1]$. This is equvalent to saying $\displaystyle xy\equiv 1(\bmod p)$. Since $\displaystyle p$ is prime it means $\displaystyle \gcd(x,p)=1$ since $\displaystyle [x]\not = [0]$. Thus, $\displaystyle ax+bp = 1$ for some integers $\displaystyle a,b$. This means that $\displaystyle ax\equiv 1(\bmod p)$ for some $\displaystyle a$. Thus, if we let $\displaystyle [y] = [a]$ then $\displaystyle [xy]=1$.
Thank you very much, perfecthacker! I totally forgot about the 'nonzero' part, so I guess I was barking up the wrong tree with that $\displaystyle xx^{-1} -1 = 0 (modp)$ stuff.

Now I'm going to be very, very cheeky and ask for any points for the 2nd bit of the proof. i.e. we can't assume p is prime, and deduce it is from the fact that $\displaystyle Z^*_p$ satisying the properties of a group.

I've outlined my thoughts in my original post, but admittedly, I don't have any concrete ideas. If you, or anyone else, can nudge me in the right direction, that'd be superb.

4. Suppose $\displaystyle n\geq 2$ (nobody ever cares for when $\displaystyle n=1$ because the group on only one element is really unintersing and never comes up anyway). We will show $\displaystyle \mathbb{Z}_n^{\text{x}}$ is not a field. If it is a field then $\displaystyle [a][b]\not = [0]$ if $\displaystyle [a],[b]\not = [0]$ (why?). But if $\displaystyle n$ is not a prime then we can write $\displaystyle n=ab$ where $\displaystyle 1<a,b<n$, thus $\displaystyle [a][b] = [n] = [0]$ while $\displaystyle [a],[b]\not = [0]$. This means it cannot be a field if $\displaystyle n$ has factors.

5. Originally Posted by ThePerfectHacker
If it is a field then $\displaystyle [a][b]\not = [0]$ if $\displaystyle [a],[b]\not = [0]$ (why?).
Thanks for the reply, but I'm confused. Consider $\displaystyle Z^*_6$.

Let $\displaystyle [a] = [2] \not = [0]$ and let $\displaystyle [b] = [3] \not = [0]$

Then surely $\displaystyle [ab] = [a][b] = (2mod6) . (3mod6) = 6mod6 = [0]$ ?

6. Originally Posted by WWTL@WHL
Thanks for the reply, but I'm confused. Consider $\displaystyle Z^*_6$.

Let $\displaystyle [a] = [2] \not = [0]$ and let $\displaystyle [b] = [3] \not = [0]$

Then surely $\displaystyle [ab] = [a][b] = (2mod6) . (3mod6) = 6mod6 = [0]$ ?
Yes [a]=[2] and [b]=[3] are non-zero and [2][3] = [6] = [0] so it means it cannot be a group under multiplication.

7. Originally Posted by ThePerfectHacker
Yes [a]=[2] and [b]=[3] are non-zero and [2][3] = [6] = [0] so it means it cannot be a group under multiplication.
D'oh.

Of course, yes.

Thank you very much for all your help.