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I have completed a proof of this that I am inclined to believe is correct, or at least on the right track. I would like to ask if it is indeed correct, or if I need a nudge in the right direction. Perhaps there is a better way to go about this?

My Proof:

(1) I will show that the existence of a generator for an infinite cyclic group implies the existence of a second which is the inverse of the former.

If $G$ is an infinite cyclic group, then every element of $G$ can be written as $a^n$ for $n\in\mathbb{Z}$ and $a$ some element of $G$. It is clear, then, that $a^{-1}$ and $a$ are both generators for $G$ since, given $n\in\mathbb{Z}$, we can choose $m\in\mathbb{Z}$ such that $(a^{-1})^n=(a)^m$ by taking $m=-n$. Moreover, we are guaranteed that $a^{-1}\ne a$ since this would imply $a=a^{-1}=e_G$ leaving us with $G$ the trivial group. So given one generator, there must be another, namely, the inverse of the given generator.

(2) Now I will show that $G$ cannot have more than two generators.

Now suppose for the sake of contradiction that $n>2$ and $G=\langle{a_1}\rangle=\ldots=\langle{a_n}\rangle$ (which is to say that any one of these $n$ elements and its inverses generate $G$), where $a_i=a_j\iff i=j$. Then, by invoking the pigeonhole principle, at least one of the members of $\{a_1,\ldots,a_n\}$ is not an inverse of one of the others. Let us call this element $a_i\ne a_1,a_{1}^{-1}$. Thus, if $a_1$ generates $G$, then there exists $b\in\mathbb{Z}$ for which $a_{1}^{b}=a_i$. Hence, if these both generate $G$, then $\forall{c\in\mathbb{Z}},\exists{j\in\mathbb{Z}}$ such that $a_{1}^{c}=a_{i}^{j}$, but $a_{i}=a_{1}^{b}$, so $a_{1}^{c}=a_{1}^{bj}$. But if $b\ne{\pm1}$, this is cannot be true, as this would state (for $c=b-1$) that $b-1=bj\implies j=1-\frac{1}{b}$ and then both $j$ and $\frac{1}{b}$ cannot be an integer for any value other than $b=\pm{1}$, which contradicts the assumption that $a_{i}$ is neither the inverse of $a_{1}$ nor equal to $a_1$.

Combining statements (1) and (2),it follows that if $G$ has one generator and it is an infinite cyclic group, then this generator is not the identity element and $G$ must have two generators. Additionally, if $G$ is an infinite cyclic group, it cannot have $n>2$ generators. These two statements mean that $G$ must have precisely two generators.

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    $\begingroup$ A bit wordy, but correct. $\endgroup$ Dec 20, 2014 at 20:13
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    $\begingroup$ if a cyclic group $G$ (finite or infinite) is generated by an element $x \in G$, then $x^{-1}$ is also a generator of $G$. so the assumption that $G$ is infinite in (1) is not necessary. $\endgroup$
    – Krish
    Dec 20, 2014 at 20:27

3 Answers 3

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You can simplify it greatly, and also do part (2) without contradiction (i.e. directly).

(1) If $G = \langle a \rangle$, then $G$ also equals $\langle a^{-1} \rangle$, because every element $a^n$ of $\langle a \rangle$ is also equal to $(a^{-1})^{-n}$.

(2) If $G = \langle a \rangle = \langle b \rangle$, then $b = a^n$ for some integer $n$, and $a = b^m$ for some integer $m$. These imply that $a = b^m = (a^n)^m = a^{nm}$; since $G$ is infinite cyclic, $nm$ must equal $1$. As an equation in the integers, $nm = 1$ has only two solutions, $(1,1)$ and $(-1,-1)$. These give $b = a$ or $b = a^{-1}$.

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  • $\begingroup$ That's a much cleaner way to go about in part (2), thanks! $\endgroup$
    – Nobody
    Dec 21, 2014 at 23:00
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One can simply use isomorphisms.

We can apply the fact that any infinite cyclic group say $G$ is isomorphic to $(\Bbb Z,+)$. Now, any isomorphism maps a generator (of the domain group) to a generator of the co-domain group, since $(\Bbb Z,+)$ has precisely $2$ generators, this implies $G$ has also $2$ generators.

PS: I concede that this is not the best solution as isomorphisms have been used, instead of concept of cyclic groups only.

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Important note: $a=a^{-1}\not\rightarrow a=\epsilon$

$a=a^{-1}\rightarrow order (a) = 2$

There are many counter-examples to this, for example all the 2-cycles in the symmetric group. However in the above scenario we are guaranteed that $a\not=a^{-1} \rightarrow \mathbf{G}$ is not infinite cyclic which is a contradiction.

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