2

I'm having difficulties with the underlined parts of the solution. enter image description here

enter image description here

usainlightning
  • 1,217
  • I don't understand why this is getting downvoted. OP is not asking for someone to do his homework, he's trying to understand the posted solution. –  Apr 09 '14 at 14:46

1 Answers1

3

1) $(\Bbb Z, +)$ is generated by $1$, so it's cyclic. The other groups have no generator; for the first, imagine we did, and say that $m/n$ generated $(\Bbb Q, +)$. But then, say, $m/2n$ would not be generated, as all elements would be of the form $km/n$, for $k \in \Bbb Z$. Similarly, pick a supposed generator of $(\Bbb Q_{>0}, \times)$. Say it's $m/n$. Pick a prime $p$ that divides neither $m$ nor $n$. Then $m^k/n^k$ can never be $p$ (why?), so this is not a generator. A similar argument works for $(\Bbb Q, \times)$.

If a group $G$ is cyclic, and $G$ is isomorphic to $H$, then $H$ is cyclic: we know that $1$ generates $G$, and our isomorphism $\varphi$ is surjective. So pick an element $h \in H$. There's some $n \in G$ such that $\varphi(n)=h$; now $\varphi(n)=n\varphi(1)$. So $h$ is a multiple of $\varphi(1)$. Since $h$ was arbitrary, $\varphi(1)$ generates the group.

2) Now we know by assumption that $\varphi(1)=f.$ Since $n \cdot (1/n) = 1$, we also know that $f = \varphi(1) = \varphi(n \cdot (1/n)) = \varphi(1/n)^n$. This implies that $f = \varphi(1/n)^n$; so $\varphi(n)$ is the $n$th root of $f$. From here, like your notes say, you can get a contradiction.