(1). How do you envisage or envision a Cyclic Group with only one generator can have at most 2 elements? Solution is based on this.

The integers have two generators, 1 and -1. Which should give you a clue as to how this is solved.

Let $G$ be a cyclic group with generator $g$. Then $g^{-1}$ is also a generator of $G$,
since any element $y$ of G can be written as $y = g^i$ for some integer $i$, and $y = \color{purple}{g^i = (g^{-1})^{-i}}$.
If G has only one generator, then $g^{-1} = g$, and therefore $g^2 = e$ = the identity element of G.
So possibility 1 is $x = e$ and G is trivial. Possibility 2 is $G = \{e, g\}$. So G has at most 2 elements.

(2.) The hinge looks like an inverse of a generator is a generator. What's the intuition? I understand the proof here on p. 7 question 11. Another proof is from $\langle a^i \rangle = \langle a^j \rangle \iff gcd(i, n) = gcd(j, n)$.
This is corollary 212 here. We just substitute into this corollary $i = 1, j = -1$.

(3.) Take a group of order $n$ and $g^i$ is a generator. What`s the intuition for $g^{ai+bn}=g^{ai} $ with $a, b$ integers? How do you intuit this when $a, b<0$? I know $ g^{ai + bn} = g^{ai}\color{darkcyan}{(g^n)}^{b} = g^{ai}{\color{darkcyan}{(e)}^b} = g^{ai} $.

  • $\begingroup$ Just saw your profile. Perhaps conjecture is another word for prognosticate. $\endgroup$ – Tyler Clark Dec 25 '13 at 9:07
  • $\begingroup$ The additive group of $\Bbb{Z}$ is just the "mother of all cyclic groups" - the others are isomorphic to its quotients, i.e. homomorphis images. Now the homomorphic image of a generator is always a generator of the image. $\endgroup$ – Jyrki Lahtonen Dec 25 '13 at 9:15
  • $\begingroup$ @TylerClark: I think we need someone like Oracle as she is in Matrix. :D $\endgroup$ – mrs Dec 25 '13 at 9:34
  • $\begingroup$ I don't understand your question in part 3 that "what's the intuition for $b$ negative?" I think maybe you are getting hung up on the idea of the negative "exponent" (if not, please let me know). Notice that if you have $g^{-b}$ this is equivalent to $(g^{-1})^b$. That is, you first take the inverse of $g$ and then take that inverse to the $b$-th power. $\endgroup$ – Tyler Clark Dec 30 '13 at 20:07
  • $\begingroup$ @TylerClark: I'm just asking about the intuition in general. But I wanted an explanation for the case $a,b < 0$. I understand the negative exponent rule but I want to intuit it here. I edited my question. Is this better? $\endgroup$ – Group Theory Jan 7 '14 at 15:19

First, what does $g^i$ mean for $i$ positive? It simply means repeated multiplication: $$g^i=\underbrace{g\cdot g\cdot\ldots\cdot g}_{\text{$i$ times}}.$$ From this, the exponent addition and multiplication rules for positive exponents follow $$\begin{aligned}g^ag^b&=g^{a+b},\\ \left(g^a\right)^b&=g^{ab}.\end{aligned}$$

We also use exponential notation for the inverse of an element $g^{-1}.$ This is an inspired notation, since it is consistent with the addition rule for exponents. For example, $$ g^3g^{-1}=(g\cdot g\cdot g)\cdot g^{-1}=g\cdot g\cdot(g\cdot g^{-1})=g\cdot g\cdot e=g^2=g^{3-1}. $$ So the $-1$ in $g^{-1}$ works like an exponent, at least in this context. We will make sure it behaves properly in other contexts in a minute.

What should $g^0$ mean? It should be defined so that the exponent addition rule works. So we need $g^ig^0=g^{i+0}=g^i.$ But that means that $g^0$ must be $e,$ since $e$ is the only group element that obeys $ae=a.$ Note that $g^0=e$ is consistent with the exponent multiplication rule as well: $\left(g^i\right)^0=e$ since any group element to the power $0$ equals $e$ by our definition. But also, by the exponent multiplication rule, $\left(g^i\right)^0=g^{i\cdot0}=g^0=e,$ which is consistent.

What should $g^{-2}$ mean? We again should define it so that the exponent addition rule holds. We have $g^2g^{-2}=g^{2+(-2)}=g^0=e.$ So $g^{-2}$ should be the inverse of $g^2.$ This again is consistent with the exponent multiplication rule: $g^{-2}=g^{2\cdot(-1)}=\left(g^2\right)^{-1},$ which is another way of saying that $g^{-2}$ is the inverse of $g^2.$ Furthermore, $g^{-2}=g^{(-1)\cdot2}=\left(g^{-1}\right)^2=g^{-1}\cdot g^{-1}.$ This is also consistent with our idea since $g^2\cdot g^{-2}=(g\cdot g)(g^{-1}\cdot g^{-1})=g\cdot g\cdot g^{-1}\cdot g^{-1}=e.$

Now for my attempt to help with your question (3): intuition behind $g^{ai+bn}=g^{ai}.$ The situation is that $g^i$ generates a cyclic group—actually a subgroup of the group generated by $g,$ where $g^n=e.$ For $g^i$ to generate a group means that the group consists of $e,$ the positive powers of $g^i,$ and the positive powers of the inverse of $g^i,$ which is $g^{-i}.$ Equivalently, the group consists of the positive, zero, and negative powers of $g^i.$

When $a$ and $b$ are positive, $g^{ai+bn}$ means $$ \underbrace{g^i\cdot\ldots\cdot g^i}_{\text{$a$ times}}\cdot\underbrace{g^n\cdot\ldots\cdot g^n}_{\text{$b$ times}}=\underbrace{g^i\cdot\ldots\cdot g^i}_{\text{$a$ times}}\cdot\underbrace{e\cdot\ldots\cdot e}_{\text{$b$ times}}=\underbrace{g^i\cdot\ldots\cdot g^i}_{\text{$a$ times}}=g^{ai}. $$ What if $a$ or $b$ is negative? Let $a=-r$ and $b=-s,$ where $r$ and $s$ are positive. Then $$ \begin{aligned} g^{ai+bn}=g^{(-r)i+(-s)n}&=\underbrace{g^{-i}\cdot\ldots\cdot g^{-i}}_{\text{$r$ times}}\cdot\underbrace{g^{-n}\cdot\ldots\cdot g^{-n}}_{\text{$s$ times}}\\ &=\underbrace{g^{-i}\cdot\ldots\cdot g^{-i}}_{\text{$r$ times}}\cdot\underbrace{e\cdot\ldots\cdot e}_{\text{$s$ times}}\\ &=\underbrace{g^{-i}\cdot\ldots\cdot g^{-i}}_{\text{$r$ times}}\\ &=g^{-ri}\\ &=g^{ai}. \end{aligned} $$

Added: I didn't address question (2) since I really wasn't sure what I could say. To my mind, the statement that the inverse of a generator is a generator is essentially part of the definition of what it means to be a generator. If $g$ generates the group, then, by definition, the group consists of the powers of $g,$ positive, zero, and negative. But since the inverse of the inverse of $g$ is $g,$ the positive, zero, and negative powers of $g$ are the same as the negative, zero, and positive powers of $g^{-1}.$ So you get the same set of elements whether you take powers of $g,$ or powers of $g^{−1}.$

I also didn't address question (1). The intuition is that, in most cases, there's more than one way to generate a cyclic group. The only exceptions are the two smallest groups, those of orders $1$ and $2.$ The get a feel for this, it helps to play with some examples. Let $C=\{e,g,g^2,\ldots,g^{n-1}\}$ be the cyclic group of order $n.$ (So $g^n=e$.) Pick some concrete value of $n,$ and have a look at what groups different elements generate. For example, if $n=6$ then the element $g^2$ generates $e,$ $g^2,$ and $g^4, $ but doesn't generate any of the odd powers. The element $g^3$ generates $e$ and $g^3,$ but no other elements. The element $g^5$ generates $e,$ $g^5,$ $g^{10}=g^4,$ $g^{15}=g^3,$ $g^{20}=g^2,$ and $g^{25}=g^1,$ which is the whole group.

As an aside, notice that for finite groups we don't get anything extra by including negative powers of a generator. For example, $\left(g^5\right)^{-1}=g^{-5}=g^1,$ $\left(g^5\right)^{-2}=g^{-10}=g^2,$ and so on, all of which are included among the positive powers. If an element has infinite order, however, its positive and negative powers are different. The reason why the negative powers give nothing new in the finite-order case is the following: let $g^n=e.$ Then if $1\le a\le n-1$, we have $g^{-a}=g^ng^{-a}=g^{n-a},$ which is a positive power. For general $a\ge1,$ we could apply this argument repeatedly until the exponent becomes positive—specifically, we could apply it at least $\lceil a/n\rceil$ times. Or we could go for complete overkill and argue that $g^{-a}=g^{na}g^{-a}=g^{na-a}=g^{(n-1)a},$ the intuition behind the decision to multiply by $g^{na}$ being that $g^{n-1}=g^{-1}$ and multiplying by $g^{na}$ will lead to a factor of $n-1$ in the final exponent. Whichever way you argue it, for every negative power there's a positive power that is equal to it.

  • $\begingroup$ Thanks a lot again. Apologies. I bungled my earlier comments. You are right. You wrote exactly what I was asking. Hence I deleted that comment. I did mean <strike>second </strike> last paragraph. I meant to ask why you wrote $g^{-a} = g^{na}g^{-a} = ...$ and not just $g^{-a} = g^{n}g^{-a} = ...$. $\endgroup$ – Group Theory Feb 4 '14 at 7:30
  • $\begingroup$ @FrankMuer: I've modified the last paragraph. The new version uses your argument and also explains what I had in mind with my argument. $\endgroup$ – Will Orrick Feb 4 '14 at 16:22
  • $\begingroup$ Thanks a lot again. Exquisite answer! I’ll upvote your other posts. $\endgroup$ – Group Theory Feb 5 '14 at 7:57

If you scroll down to the other answer given, the person says that $\mathbb{Z}$ has only one generator and infinitely many elements. The first paragraph of the answer you are referring to is just saying that $\mathbb{Z}$ actually has two generators; ergo, it is not a counter example to the question.

The second paragraph of the answer you are referring to gives the complete proof of the problem. You can see #1 in http://math.berkeley.edu/~ogus/old/Math_113/sol3.pdf for reference. You don't even need to concern yourself with $g^i= (g^{-1})^{-i}$.

  • $\begingroup$ Thanks. I upvoted. I updated my question because I realized what really was baffling me. Can you please look at it? Are you referring to 1 in your linked pdf? $\endgroup$ – Group Theory Dec 30 '13 at 13:40

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