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I have to find generators for the group $(A,*^._9)$ where $A =${$2,4,8,7,5,1$}.

The result to the problem is 2, which creates {$2,4,8,7,5,1$}.

Initially, I though the solution was 1, as it creates the exact same list, but this seems to be wrong.

I don't quite understand... Why is it that 2 is a generator but 1 isn't although they generate the exact same list?

Talha Ahmed
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Johny
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3 Answers3

5

Assuming that $A$ is the multiplicative group of the integers mod $9$, note that $1$ is not a generator as $1^n\equiv1\pmod9$ for all $n\in\mathbb Z^+$.

Now $4^3=64\equiv1\pmod9$ so $4$ is not a generator. Similarly $7^3=343\equiv1\pmod9$ and $8^2=64\equiv1\pmod9$ so $7$ and $8$ are not generators.

On the other hand, you should find that $2,2^2,\ldots,2^6$ is the set $\{1,2,4,5,7,8\}$ mod $9$; similarly for $5,5^2,\ldots,5^6$, hence $2$ and $5$ are generators.

Olinguito
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I'm back-interpreting some of your notation, as the question is imprecise at the moment. But I think that having a reasoned-out answer would be helpful.

Firstly, a group $G$ is (roughly) a set of elements along with an associative group law or way of combining two elements $g,h$ of $G$ to get another element $gh$ of $G$, satisfying some properties.

So the first question we should ask is, what is your group, and what is the group law? You mention that your group's elements are $\{2,4,8,7,5,1\} = \{1,2,4,5,7,8\}$, and the group multiplication law is $*_9$. Ok, well, what is $*_9$?

I would guess that $*_9$ means multiplication mod $9$. So given two elements $g,h$ in $A$, we interpret $gh$ to mean the number we get after multiplying $g$ and $h$ modulo $9$.

Looking at powers of $2$, we see that $2 = 2, 2^2 = 4, 2^3 = 8, 2^4 = 16 = 7, 2^5 = 32 = 5, 2^6 = 64 = 1$, and these numbers $2,4,8,7,5,1$ are exactly the elements of your group. So $2$ is a generator.

You conjecture that $1$ is also a generator. Let's check --- what are the powers of $1$ mod $9$? They are $1, 1^2 = 1, 1^3 = 1, \ldots$ Thinking about it, since our group law is multiplication mod $9$, of course $1$ can't be a generator. When raising $1$ to powers, we only get $1$.

However, there are more generators. For instance, $5$ is a generator. The powers of $5$ are $5 = 5, 5^2 = 7, 5^3 = 35 = 8, 5^4 = 40 = 4, 5^5 = 20 = 2, 5^6 = 10 = 1$. All the elements of $A$ appear in this list as well.

davidlowryduda
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  • That cleared it up very well, thank you. – Johny Dec 23 '16 at 18:46
  • One thing I am missing is where do we define that we want to multiply the generator to the power of the number of elements in the list. – Johny Dec 23 '16 at 18:59
  • @Johny I stopped when we got to $1$, as it then repeats if you continue to take higher powers. This will always happen eventually (in a finite group), and an element is a generator exactly when all values in the group are taken. – davidlowryduda Dec 24 '16 at 15:16
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You seems to be considering the multiplicative group $(\mathbf{Z}/9\mathbf{Z})^\times$ of units of the additive group $\mathbf{Z}/9\mathbf{Z}$. The whole difficulty is to understand the fundamental difference between the easy cyclic structure of $\mathbf{Z}/9\mathbf{Z}$, unlike the more tricky one of $(\mathbf{Z}/9\mathbf{Z})^\times$.

$1$ does not generate $A$, for you are talking about the multiplicative group, and not the ring structure of $\mathbf{Z}/9\mathbf{Z}$. The group generated by $1$ is all the powers of $1$, that is the trivial subgroup $\{1\}$. The group generated by $2$ is all the powers of $2$, and this explains the order you wrote the numbers in the set, for modulo 9 you get:

$$2^1=2$$ $$2^2=4$$ $$2^3=8$$ $$2^4=16=7$$ $$2^5=32=5$$ $$2^6=64=1$$

Hence $2$ is indeed a generator.

Desiderius Severus
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