So, I'm relearning Group Theory. And I got the axioms down, I think. So let's make a concrete example:

  • The collection of numbers the positive integers less than 7: 1,2,3,4,5,6
  • The • operation will be multiplication mod 7.
  • Associativity holds.
  • The Identity e is 1.
  • Every element has an inverse:
    • 1*? mod 7 = 1 --> 1
    • 2*? mod 7 = 1 --> 4
    • 3*? mod 7 = 1 --> 5
    • 4*? mod 7 = 1 --> 2
    • 5*? mod 7 = 1 --> 3
    • 6*? mod 7 = 1 --> 6

But! What is the order of the group?! I thought the order would be 7. But there are 6 elements! So maybe I was wrong and 0 should be in the group.

But 0 does not have an inverse! There is no x such that 0*x mod 7 = 1.

So what am I misunderstanding here? Is it the definition of order? Is it some other trick about groups?


The only error is your belief that the order "should" be 7. The order of a finite group is just the number of elements in the group. Your group consists of the positive integers that are smaller than, and relatively prime to, 7. There are six of them, so your group has order 6.

(I'm not sure why you thought the order should be 7...)

Indeed, you cannot add $0$ to the mix and still have a group. If you consider the numbers $0,1,\ldots,6$ under multiplication modulo $7$ you do not get a group, you get a semigroup.

Added: Ah, Jonas Meyer's reply suggests what is going on; since you say you are relearning Group Theory, you might have vague memories of the "group of integers modulo $n$" as having order $n$. The group of integers modulo $n$ under addition has order $n$; but the multiplicative group modulo $n$ consists of the positive integers less than, and relatively prime to, $n$, with the operation being multiplication modulo $n$, and has $\varphi(n)$ elements (Euler's phi function). When $n=7$ (the case you are looking at), the group has $\varphi(7)=6$ elements, as you observed.

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    $\begingroup$ For any ring R the set of units (invertible elements) comprise a group U(R) called the unit group of R. $\endgroup$ – Bill Dubuque Nov 6 '10 at 3:17

You're right, the group has order 6 because it has six elements. You can make {0,1,2,3,4,5,6} a group with addition mod 7. This would be a group of order 7.


Good thinking. 0 is not in the group, so the order is 6. 0 is in another group in the multiplicative $\mathbb{Z}/\mathbb{Z}_7$, which is the trivial group (one element). The groups are disjoint.


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