If one group is cyclic and the other isn't, can they be isomorphic?
So I looked through the isomorphic properties and they seem to satisfy all of them...Prove that additive groups and are not isomorphic
- The order of equals the order of
- Both are abelian
- The order of any element in the groups are all infinite (except the identity)
The only thing I can think of is that has a generator and doesn't, but does that have anything to do with them being isomorphisms?
As noted above, we need to prove $\Bbb Q$ is NOT cyclic, which would settle the question (since an isomorph of the integers would be cyclic).
So, suppose $\Bbb Q$ had a generator, which we will write as $q = \dfrac{a}{b}$ with $\text{gcd}(a,b) = 1$ (a fraction "in simplest terms"), and of course $b \neq 0$. We can further insist that $b > 0$, for if not, replace $\dfrac{a}{b}$ with $\dfrac{-a}{-b}$.
Consider $r = \dfrac{1}{b+1}$, which is clearly an element of $\Bbb Q$. Since we are supposing $\Bbb Q$ cyclic, there is some INTEGER $k$ with:
$k\left(\dfrac{a}{b}\right) = \dfrac{1}{b+1}$.
This leads to the equation of integers:
$ka(b+1) = b$, which after some re-arranging, we can write as:
$k = b(1 - ka)$.
It is clear from this equation that $b$ divides $k$, so we have $k = nb$ for some integer $n$. This means that $k\left(\dfrac{a}{b}\right) = \dfrac{ka}{b} = \dfrac{nab}{b} = na \in \Bbb Z$.
Thus:
$na = \dfrac{1}{b+1}$, so that $na(b+1) = 1$. Thus $b+1$ divides 1, and since $b+1 > 0$, we must have $b+1 = 1$ so that $b = 0$, contradiction.
In short, for any element $\dfrac{a}{b}$, we have found an element $r$ NOT in $\left\langle \dfrac{a}{b} \right\rangle$, namely $r = \dfrac{1}{b+1}$.