Prove that $\mathbb{Z}_n$ has an even number of generator for n>2

Question

My proof of this question is as following

Let $x$ be the element of the generator for the cyclic group $\mathbb{Z}_n$, so that that the order of $x$ is $n$, i.e. $x^n=x^{n-1}\circ x=x\circ x^{n-1}=e$, so $x^{n-1}$ must be the inverse of $x$ in the group. Then because $\gcd(n,n-1)=1$, then the order of $x^{n-1}$ is $n/1=n$. Therefore $x^{n-1}$ is a generator of the group $\mathbb{Z}_n$, since $<x>=<x^{n-1}>$. Hence for every element $x$ that is the generator of the group $\mathbb{Z}_n$, we can always have its inverse element $x^{n-1}$ or $x^{-1}$ as another generator of the group. i.e. if I find odd numbers of generator for the group, I must have an even number of generators, or if I find even number of generators, I must have an even number of generators as well.

I don't know if my proof is correct enough so I want to see what your guys thinking. And perhaps if it is possible, I have a second question which I don't know how exactly to prove. here is the statement

A group with a finite number of subgroups is finite.

I intuitively suggest this must be true but I can't write reasonable proof for it.

Thanks.

What's this cyclic group $\Bbb Z_n$ with a multiplicative law? — Bernard, Oct 03 '20 at 12:16
You should probably say how you know that $x$ and it’s inverse are not equal to each other. Also, since you’re working in an abelian group whose operation is denoted additively, it might be easier to read if you use $-x$, $nx$ , etc rather than the abstract multiplicative notation. — Ned, Oct 03 '20 at 12:19
Using that the number of generators is $\varphi(n)$ is an easier approach. — Peter, Oct 03 '20 at 12:20
For the second question, see https://math.stackexchange.com/q/22996/62967 — egreg, Oct 03 '20 at 12:28
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score 4 · Accepted Answer · answered Oct 03 '20 at 12:32

4

More abstract: suppose that the group $G$ is cyclic; then, for each generator $x$, also $x^{-1}$ is a generator, because $x^k=(x^{-1})^{-k}$.

Suppose $x=x^{-1}$; then $x^2=1$ (or $e$, if you prefer this notation; I don't) and therefore $|G|\le2$.

Thus, if $|G|>2$, we have $x\ne x^{-1}$, for every generator $x$, and thus we can divide the generators into pairs.

Hint for the second question: any group is the union of its cyclic subgroups; if the number of subgroups is finite, none of the cyclic subgroups can be infinite.

answered Oct 03 '20 at 12:32

egreg

244,946

Thank you. I can't exactly catch what you mean by that if the number of subgroups is finite, none of the cyclic subgroups can be infinite. – oscarmetal break Oct 03 '20 at 12:39
I mean how you connect the idea between number of subgroup, and the order of a cyclic subgroup() – oscarmetal break Oct 03 '20 at 12:47
@oscarmetalbreak If a cyclic group is infinite, it has infinitely many subgroups (it is isomorphic to the integers); the existence of only finitely many subgroups rules this case out. Thus we have a finite union of finite cyclic groups. – egreg Oct 03 '20 at 13:40
Please tell me if I understand it correctly. Because a group is a union of its cyclic subgroups. So finite number of subgroups indicates finite number of cyclic groups. And the cyclic group must be of finite order, since infinite cyclic groups produce infinite number of subgroups. Hence a group being its finite number of cyclic subgroup of finite order must be a finite group. – oscarmetal break Oct 03 '20 at 14:30
@oscarmetalbreak That's it. – egreg Oct 03 '20 at 14:41

score 1 · Answer 2 · answered Oct 03 '20 at 12:25

1

I like it. Alternatively, the number of generators of $\Bbb Z_n$ is $\varphi(n)$, where $\varphi$ is Euler's totient function. A basic fact about $\varphi$ is that it's values are always even.

For your second question, the hint is to look at cyclic subgroups, noting that the infinite cyclic subgroup, $\Bbb Z$, has infinitely many subgroups.

answered Oct 03 '20 at 12:25

We have not been taught the Euler's totient function yet so I don't know how to use $\varphi$. – oscarmetal break Oct 03 '20 at 12:31
Oh well. You will meet it. – Oct 03 '20 at 12:35

Alfha · Answer 3 · 2022-02-12T05:03:10.783

Let $\mathbb{Z}_n$ be cyclic group of order $n \gt 2$ then either $n = 2^s$ such that $s \gt 1$, say if $s = 1$ then $n = 2^1 = 2$ but $n \gt 2$ so $s$ has to be more $1$. Then no. of generators of $\mathbb{Z}_n$ = no. of elements of order $n=$ $$ \varphi(n) = \varphi(2^s) = 2^s - 2^{s-1} = 2^{s-1}(2-1) = 2^{s-1}$$ which is divisible by $2$ for $s$ more than 1, so $\varphi(n)$ is even.

If $n$ is not of the form $2^s$ then $n$ must be of the form $p^kt$ such that $p$ is prime greater than $2$, $p$ and $t$ are co-prime and $n$ is $t$ multiple of $p^k$ then (Can you guess further steps of the proof?)

Spoiler $\varphi(n) = \varphi(p^kt)$ $= \varphi(p^k)* \varphi(t)$ $= (p^k - p^{k-1})* \varphi(t)$ $= p^{k-1}(p-1)* \varphi(t)$. Note that $p$ is prime more than $2$ then $p$ must be odd and $p-1$ must be even and as $\varphi(n)$ is multiple of $(p-1)$ then $\varphi(n)$ must be even.

Hence $\mathbb{Z}_n$ must have even no. of generators if $n$ is more than 2. Infact you can generalize above result for any cyclic group with order more than $2$.

Prove that $\mathbb{Z}_n$ has an even number of generator for n>2

3 Answers3