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These statements are false according to my book. I am not sure why though

  1. In every cyclic group, every element is a generator
  2. A cyclic group has a unique generator.

Both statements seem to be opposites.

I tried to give a counterexample

  1. I think it's because $\mathbb{Z}_4$ for example has generators $\langle 1\rangle$ and $\langle3\rangle$, but 2 or 0 isn't a generator.

  2. As shown in (1), we have two different generators, $\langle1\rangle$ and $\langle3\rangle$

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You have answered the question fully. –  André Nicolas Oct 16 '12 at 1:53
    
What about something like $\pi^n$ where n is integer? That was my first attempt, but I wasn't 100% sure. I thought that for some reason $1/\pi$ can't be written (generated) by $\pi^2$. NBut all I have to do is multiply by -1 on the two right? –  sidht Oct 16 '12 at 1:59
    
The powers of $\pi$ are isomorphic to the additive group $\mathbb{Z}$. Cetainly $\pi^2$ is not a generator, we just get the even powers of $\pi$. The non-negative powers are too big, and the negative powers too small, to yield $\frac{1}{\pi}$.So indeed not every element is a generator. And there is not a unique generator, since $\pi$ and $\frac{1}{\pi}$ are both generators. Thus that is also a counterexample to the assertions. –  André Nicolas Oct 16 '12 at 2:18
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2 Answers 2

up vote 1 down vote accepted

Your examples work nicely. Let $\mathbb{Z}_n$ be the cyclic group of order $n$. The following theorem is useful in looking at this sort of situation (taken from Contemporary Abstract Algebra by Gallian, 5th ed.):

Let $a$ be an element of order $n$ in a group and let $k$ be a positive integer. Then $\langle a^k\rangle=\langle a^{\gcd(n,k)} \rangle$ and $|a^k|=n/\gcd(n,k)$.

So how can we apply this? Well, clearly $1$ is an element of $\mathbb{Z}_n$ of order $n$. Then (now in additive notation to be consistent with the operation in $\mathbb{Z}_n$) $\langle k\cdot 1 \rangle = \langle k \rangle = \langle \gcd(n,k)\cdot 1 \rangle$, and so if $\gcd(n,k)=1$ then $\langle k \rangle = \langle 1 \rangle = \mathbb{Z}_n$.

This also follows by the second part of the theorem, by noting that $|\langle k \rangle |=|k|$. If $\gcd(n,k)=1$ then $|k|=\frac{n}{1}=n=|\langle k \rangle |$ so $k$ generates $\mathbb{Z}_n$. On the other hand, if $m=\gcd(n,k)\ne 1$ then $|k|=\frac{n}{m}=|\langle k \rangle |<n$, so $k$ does not generate $\mathbb{Z}_n$ but rather a subgroup of order $\frac{n}{m}$.

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Take $Z_n$. This group is cyclic and the generators are $\phi(n)$ = all the numbers that are relatively prime to $n$

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