# Two small group questions

1. Feb 18, 2013

### CAF123

1. The problem statement, all variables and given/known data
1)Fix $n \in \mathbb{N}$. Consider multiplication mod $n$. Let G be the subset of {1,2,...,n-1} = $\mathbb{Z}_n$ \ $\left\{0\right\}$ consisting of all those elements that have a multiplicative inverse (under multiplication mod n). Show that G is a group under multiplication. Describe the group when n =12.

2) Show that if G is cyclic, then G is abelian.
3. The attempt at a solution

1) The reason I can't make much progress here is essentially interpretation. '..that have a multiplicative inverse (under multiplication mod n)' What does this mean?
What I said intially was multiply a number, p, in {1,2...n-1} by x and then take that number mod n and if there exists an x such that this equals 1, then that is in G. However, considering some cases, n=3,4 etc.. there exists an x for all elements. I tried something else that made me get a single element for G for all n. What is the appriopriate interpretation?

2)I know the proof to this since it is in my textbook, but I tried it slightly different and I want to know if it is valid:
Consider two elements, $g_1, g_2$, with $g_1$ the generator of the group and $g_2$ some element generated by $g_1$. Then consider the case of $g_1 = g_2$. Since G is cyclic, this means, $g_2 = g_1^n = g_1$, n the order of $g_1$.
Now multiply by $g_2$ on both sides: $g_1 g_2 = g_2 g_2 = g_2 g_1^n = g_2 g_1$.

Many thanks.

2. Feb 18, 2013

### Dick

For 1) you already have the correct interpretation. What did you get for the inverse of 3 mod 12? I don't think every element has an inverse. For 2) I'd stick with the books proof. 'Multiplying by g2 on both sides' doesn't make much sense when you are trying to prove the group is abelian.

3. Feb 18, 2013

### CAF123

Take for example n= 3 so the set is {1,2}. Then:$$(1 \cdot x )mod 3 =1 \Rightarrow x = 1 \,\,\text{and}\,\,( 2 \cdot x) mod 3 = 1 \Rightarrow x =2$$ so G ={1,2}.

Similarly, for n=12, if p=1, x=1, if p=2 x does not exist ... (there is no x when multiplied by 2 and then taken modulo 12 will give 1) I see...

What do you mean it doesn't make sense? All I was doing was taking another element from the group and smashing it with another element on the right side and then using properties of cyclic groups to establish the abelian result. Am I not allowed to do this?

Many thanks

4. Feb 18, 2013

### CAF123

1)After computing for some values of n, I think I can see that if a number divides n, then it is not in G.
Closure: Take g1, g2 in G, then (g1g2)mod n is in G, by the above observation.
Associativity: $g*(h*k) = g_1 * (g_2g_3 mod n) = g_1 * ([g_2 mod n][g_3 mod n] mod n) = (g_1 mod n)(g_2 mod n)(g_3 mod n) = (g*h)*k.$
Identity: Take g in G = (mn + 1) mod n, where m is a scalar multiple of n
Inverse: By defintion of how the group G was constructed, there exists an inverse element to all those elements in G.

Is it ok?

5. Feb 18, 2013

### Dick

I mean it doesn't make sense. I do agree that if g1=g2, then g1*g2=g2*g1=g1*g1=g2*g2. But that's true in any group abelian or not. That's really all I see in your argument.

6. Feb 18, 2013

### CAF123

Up to what you have written is indeed true for any group, but then I take it further and say because G is cyclic, I can express g2 in an alternate way and thus arrive at the conclusion of g1g2 = g2g1. No?

BTW I am not saying I am definitely right, I just want to make sure I understand where the fault is. Thanks.

7. Feb 18, 2013

### Dick

If a number divides n then it is not in G, true. But it's also true that 9 is not in G if G is the integers mod 12. If a number has any common divisor with n, then it's not in G. The name of the condition you want is 'relatively prime'. And rethink some of those arguments, start with the last one. How about just showing 1 is the identity?

8. Feb 18, 2013

### CAF123

What ones were not clear?

For the identity, g = (mn +1) mod n = 1 = e.

9. Feb 18, 2013

### Dick

Start with the first one. Since 'doesn't divide n' is not the correct criterion for having an inverse mod n, 'by the above observation' doesn't really prove much.

10. Feb 18, 2013

### CAF123

Take g1, g2 in G. Then g1g2 will also not have a common divisor with n. (Since if it did, one or both of g1 or g2 would have to have a common divisor with n). So, on division with n, we have some remainder, r < n in G. (Closure)

Associativity: g1modn ((g2modn)(g3modn)) = (g1 +an)(g2 + bn)(g3 +cn), a,b,c in N. Since these are just numbers, they obey a(bc) = (ab)c.

Identity: As before, take g = (mn +1)mod n = 1.

Inverse. (g1 mod n) . x = 1 => ( g1 + an) .x = 1 => x = 1/(g1 + an). I don't think I have to show this because the only reason I have g1 is because by defintion it has a multiplicative inverse under mod n.

PS: Did you see my Post #6?

Many thanks.

11. Feb 18, 2013

### Dick

Those are looking better. And, yes, you don't have to prove the elements have inverses since you are only picking the ones that do. A more direct way to prove closure is just that if a and b have inverses, then the inverse of ab is b^(-1)a^(-1). It doesn't really depend on some special property of the group.

Yes, I did see you post 6. If you are talking about the proof a cyclic group is abelian, maybe you have some kind of idea, but your proof doesn't convey that to me. I just see something saying if g1=g2, then they commute.

Last edited: Feb 18, 2013
12. Feb 18, 2013

### CAF123

So essentially I have only proved the case where g1 = g2 and not proved the general case for arbritary g? I see the fault now,thanks.