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I've been reviewing some abstract algebra for an upcoming midterm, and I seem to be stuck on this part:

Is it possible for an infinite, non-cyclic group to exist?

I thought maybe the group of all integers is an infinite non-cyclic group, but I'm not sure if that's true. Any input would be really helpful, as I'm very shaky on cyclic/non-cyclic groups. Thanks.

Max
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A cyclic group is a group that is generated by a single element, i.e. it is of the form $\{a^n : n\in\Bbb Z\}$. The group of integers with addition satisfies this, as it is the group of multiples of $a=1$. But note that a cyclic group is necessarily abelian (the powers of an element commute with each other). So any non-abelian infinite group would satisfy your requirement. Take for example the group of invertible square matrices of a given size over some ring.

zarathustra
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  • Oh, so since matrices aren't always commutative, this implies that any group of MxN matrices are non-abelian and therefore form an infinite, non-cyclic group? – Max Sep 18 '16 at 07:24
  • Exactly! (Note that $m\times n$ matrices do not form a group unless $m=n$.) – zarathustra Sep 18 '16 at 07:25
  • "this implies that any group of MxN matrices are non-abelian" -- no, not any group of matrices. Take the invertible diagonal matrices of size $n$. They form an abelian group. Usual $n\times n$ matrices do not form a group under multiplication, since not every matrix is invertible. – Dietrich Burde Sep 18 '16 at 13:39
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The group of integers is indeed cyclic: $\mathbb{Z}=\langle 1\rangle$ because $n=1+1+\dots +1$ $n$ times if $n\ge 0$ and $n=(-1)+(-1)+\dots +(-1)$ $-n$ times if $n<0$.

Infinite non-cyclic groups do exists. An easy example is the abelian group $\mathbb{Z}_2\times\mathbb{Z_2}\times\dots$ because any element in it has order $2$.

Another example is $\mathbb{Q}$. You can show that for any $p,q\in\mathbb{N}$ you can find a rational number $r$ such that $r\notin\langle \frac{p}{q}\rangle$. You can find a proof here.

I give as a last example the infinite dihedral group. It is defined as $D_\infty=\langle r,s\mid s^2=e,\,rs=sr^{-1}\rangle$. Note that the order pf $r$ is infinite. This group can be seen to act on $\mathbb{Z}$ as follows: $r$ is translation by $1$, to the right for example (you can choose to the left, then $r^{-1}$ translates to the right), and $s$ is reflection in the origin: any integer switches its place with its opposite.

Community
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Scientifica
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Most infinite groups are not cyclic, because they are not even abelian. An important class here is given by matrix groups over an infinite domain, e.g., by $$ GL_n(K),\; SL_n(K) $$ for $n>1$ and an infinite field $K$; or $GL_n(\mathbb{Z}),\; SL_n(\mathbb{Z})$ over the integers. In particular, all free groups $F_n$ of rank $n>1$ are infinite, non-cyclic, because they can be embedded into $SL_2(\mathbb{Z})$.

Dietrich Burde
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The simplest infinite non-cyclic abelian groups: $\mathbf Z^n$, with $n>1$.

They are non-cyclic, because they are $\mathbf Z$-modules, have a basis of $n$ elements, and commutative rings have the *invariant basis number$ property.

Bernard
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