1
$\begingroup$

Show that $G=\{a+b\sqrt2\mid a,b\in\mathbb{Z}\}$ is not cyclic under addition.

Trying to show that $G$ is not a group, and without showing that $G$ is isomorphic to $\Bbb{Z}\times\Bbb{Z}$, it seems somewhat strange to show that $G$ cannot be generated by an element of $G$

Edit: I realized I left off some pertinent information. I am supposed to show it is not cyclic.

$\endgroup$
  • 2
    $\begingroup$ Several misconceptions at work here: 1. G is a group. 2. G has a closure (!). 3. You are asked to show that G is not equal to its closure. 4. The usual way to do this is to show that G is dense. Can you do that? $\endgroup$ – Did Feb 7 at 12:27
  • $\begingroup$ $G$ is an additive subgroup of the real numbers. $\endgroup$ – ncmathsadist Feb 7 at 12:50
  • $\begingroup$ The "closure" here is most likely the closure as a subspace of the (topological or metric) space $\Bbb R$, not some sort of algebraic or arithmetic closure. $\endgroup$ – Arthur Feb 7 at 13:11
  • $\begingroup$ @Jac Frall About which operation we say? $\endgroup$ – Michael Rozenberg Feb 7 at 13:26
  • $\begingroup$ @MichaelRozenberg under addition $\endgroup$ – Jac Frall Feb 7 at 14:02
0
$\begingroup$

Suppose it were cyclic. Then, there exists a $c \in G$ such that every $d \in G$ is of the form $cn$ for some $n \in \mathbb Z$.

Now, note that $1 = cm$ for some $0 \neq m \in \mathbb Z$ and $\sqrt 2 = cn$ for some $0 \neq n \in \mathbb Z$. Therefore, $c$ is rational, as $c = \frac 1m$, but then $\sqrt{2} = \frac nm$ is rational, a contradiction.


Alternately, you may also generalize this idea to the following result :

Let $\{1,m_1,...,m_n\} \subset \mathbb R$ be linearly independent over $\mathbb Q$ (that is, if $a_0,a_1,...,a_n \in \mathbb Q$ are such that $a_0+\sum_{i=1}^n a_im_i = 0$ then $a_i = 0$ for all $i$), with $n \geq 1$. Then, the group $$G' = \mathbb Z[m_1,...m_n] := \left\{\sum a_im_i : a_i \in \mathbb Z\right\}$$ cannot be cyclic.

It is a more difficult exercise to show that the above cannot be generated even by any $n-1$ elements over $\mathbb Z$.

$\endgroup$
  • $\begingroup$ Exactly what I was looking for. $\endgroup$ – Jac Frall Feb 8 at 18:50
1
$\begingroup$

Hint : The group G is generated by the set $\{1, \sqrt{2} \}. $ Cyclic group is generated by single element of the group .

Or

Let if possible group is cyclic then $G=c\mathbb{Z}$ which implies that $c$ is irrational number and $G=c\mathbb{Z}$, a contradiction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.