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(i) Let $G = \langle a \rangle$ have order $rs$, where $(r,s) = 1$. Show that there are unique $b,c \in G$ with $b$ of order $r$, $c$ of order $s$ and $a = bc$.

(ii) Use part (i) to prove that if $(r,s) = 1$, then $\phi(rs) = \phi(r) \phi(s)$.

I would like to a hint. I don't have idea how (i) helps me to prove (ii).

P.S.: $\phi$ is the Euler function: $\phi(1) = 1$ and $\phi(n) = |\{ k \ ; \ 1 \leq k < n \ and \ (k,n) = 1 \}|$.

EDIT:

My attempt after receive the hint.

Affirmation 1: Let be $G = \langle a \rangle$ with order $n$, then $a^k$ is a generator of $G$ if and only if $(k,n) = 1$

$(\Longrightarrow)$

Exists $x \in \mathbb{Z}$ such that $(a^k)^x = a$, then $kx \equiv 1 \mod n$ and then exists $y \in \mathbb{Z}$ such that $kx + ny = 1$, so $(k,n) = 1$

$(\Longleftarrow)$

exists $x, y \in \mathbb{Z}$ such that $kx + ny = 1$, then

$$a^{kx + ny} = a$$ $$a^{mkx+mny} = a^m$$ $$a^{mkx} = a^m$$ $$(a^k)^{mx} = a^m, \forall m \in \mathbb{Z}. \square$$

By the part $(i)$ of exercise, $|\langle a \rangle| = |\langle b \rangle| |\langle c \rangle|$ and $\phi(rs) = |\langle a \rangle|$, $\phi(r) = |\langle b \rangle|$ and $\phi(s) = |\langle c \rangle|$ by the Affirmation 1, then $\phi(rs) = \phi(r)\phi(s)$. $\square$

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George
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Hint: If $G$ is a cyclic group of order $rs$, then $\phi(rs)$ is exactly the number of elements of $G$ which generate $G$ (why?). So you want to show such a group has $\phi(r)\phi(s)$ different generators. Part (i) says that each such generator has the form $bc$ for unique $b,c\in G$ of order $r$ and $s$. How many such $b$ and $c$ are there?

Eric Wofsey
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  • I edited my post and put my attempt. Is it right? – George Mar 01 '17 at 00:07
  • Your argument for Affirmation 1 is good. The final line after that seems to not quite be written correctly. For instance, every time you write $\langle a\rangle$ you seem to instead mean the set of all elements which generate $G$. Also, Affirmation 1 directly gives that this set has $\phi(rs)$ elements, but you need to say a little more to explain why the sets of possible values of $b$ and $c$ have the sizes you claim. – Eric Wofsey Mar 01 '17 at 00:15
  • Sorry, but I didn't understand why Affirmation 1 doesn't ensure that $|\langle b \rangle| = \phi(r)$ and $|\langle c \rangle| = \phi(s)$. Can you explain me why I can't affirm that $|\langle x \rangle| = \phi(m)$ for every cycle $\langle x \rangle$ with order $m$? – George Mar 01 '17 at 17:52
  • First of all, your notation is still wrong: $|\langle x\rangle|$ is not the number you want to be referring to. Second, Affirmation 1 tells you that $\langle x\rangle$ has $\phi(m)$ elements which generate all of $\langle x\rangle$ (that is, $\phi(m)$ elements of order $m$). But you want to count the number of elements of order $r$ (or $s$) in the entire group $G$, not just in $\langle b\rangle$ (or $\langle c\rangle$) for some fixed $b$ (or $c$). So you need to explain why these are the same thing. – Eric Wofsey Mar 01 '17 at 17:57
  • @EricWofsey If $(r,s)=1$, then $\langle a^{pr+qs} \rangle = \langle a^{pr} \rangle \langle ^{qs} \rangle$. This shows that $\phi(rs) = \phi(r)\phi(s)$? – Blue Tomato Mar 10 '23 at 21:28