1

I have to prove that $\mathbb{Z}/n\mathbb{Z}$ is not a group under multiplication for all n>1. I would argue, however, that when $n$ is prime, $\mathbb{Z}/n\mathbb{Z}$ is a group under multiplication. Multiplication is associative, every element in $\mathbb{Z}/n\mathbb{Z}$ has an inverse when $n$ is prime, and the identity element is simply $\bar{1}$. What am I missing?

Obliv
  • 407
  • 3
    $0$ hasn't got an inverse. – lulu May 13 '16 at 17:50
  • oh. So then isn't it true that $\mathbb{Z}/n\mathbb{Z}$ is not a group under multiplication for all $n$? – Obliv May 13 '16 at 17:51
  • 1
    As lulu noted, $0$ has no inverse. Though, $\Bbb Z/p\Bbb Z \setminus {0}$ is a group –  May 13 '16 at 17:51
  • 1
    @Obliv That's a lot of negatives. $\mathbb Z / n\mathbb Z$ is never a group under multiplication because the element $0$ is never invertible. – lulu May 13 '16 at 17:52
  • how do I prove,then, that $\mathbb{Z}/n\mathbb{Z}$ is not a group under multiplication? Do I simply show how 0 is not invertible and that 0 is in $\mathbb{Z}/n\mathbb{Z}$ for all n>1? – Obliv May 13 '16 at 17:54
  • That looks like a proof to me. We know $0$ is not invertible because $a\times 0=0;\forall a\in \mathbb Z$. – lulu May 13 '16 at 17:56
  • I appreciate your help. Thank you :)! – Obliv May 13 '16 at 17:58
  • Z/nZ is never a group Z/pZ-{0} is if p is prime. Z/nmZ-{0} is not. – fleablood May 13 '16 at 18:28

3 Answers3

3

For $\mathbb{Z}/n\mathbb{Z}$ to be a group, we need that for every $a \in \mathbb{Z}/n\mathbb{Z}$, to exist $b \in \mathbb{Z}/n\mathbb{Z}$ such that $a*b = 1$. But for $a = 0$, there is no $b$ such that $a*b = 1$, because $a*b = 0, \forall b \in \mathbb{Z}/n\mathbb{Z}$. Then, $\mathbb{Z}/n\mathbb{Z}$ is not a group!

Luísa Borsato
  • 2,316
1

Maybe it worth asking when is $(Z/nZ)^*$ a group under multiplication and the answer is precisely when $n$ is prime. Otherwise, $n=ab$ would imply that the product of the residue classes of $a$ and $b$ does not belong to $(Z/nZ)^*$

joy
  • 440
  • 1
    yeah but that doesn't change the fact that $0$ is in $\mathbb{Z}/n\mathbb{Z}$ regardless of what $n$ is. It'll never be a group as long as $0$ is in it. – Obliv May 13 '16 at 18:06
  • True. That's why the question is not interesting. But the absence of $0$ makes things interesting. – joy May 13 '16 at 18:11
  • The question may not be interesting but it was the question that was asked. No amount of interesting is going to outweigh relevence, I'm afraid. It's precisely your interesting case, that caused the OP trouble in the first place so reiterating it makes things worse. Not to mention, the set you are talking about is NOT Z/nZ but (Z/n) - {0}. That detail is extremely important. – fleablood May 13 '16 at 18:21
  • By adding just a remark that concerns the real question of the OP you could make your answer complete. – drhab May 13 '16 at 18:50
0

No. But it is an cyclic additive group for all $n$. And for each $n$ , you can consider the multiplicative group of units, $\Bbb Z_n^×$, which, for $p$ prime coincides with $\Bbb Z_p^*$, the multiplicative group of nonzero elements.