If $(r,s)=1$, then $\phi(rs)=\phi(r)\phi(s)$

Question

It is required to show that if $(r,s)=1$, then $\phi(rs)=\phi(r)\phi(s)$, with an argument of group theory, where $\phi$ is the Euler totient function and $(\cdot,\cdot)$ is the gcd.

This is the second exercise of the following problems:

$(i)$ 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 $\phi(rs)=\phi(r)\phi(s)$.

Number $(i)$ was solved without problem. Note that $b=a^{qs}$ and $c=a^{pr}$ were chosen.

Number $(ii)$ is not clear the argument. Here is what has been tried:

Let $(r,s)=1$. By $(i)$, it follows that $\langle a \rangle = \langle a^{pr}a^{qs} \rangle$ for some integers $p,q$.

Claim: $\langle a^{pr} \rangle \langle a^{qs} \rangle = \langle a^{pr}a^{qs} \rangle$.

Let $x \in \langle a^{pr} \rangle \langle a^{qs} \rangle$, then $x = (a^{pr})^{k_1} (a^{qs})^{k_2}$ for some integers $k_1, k_2$. Thus, $x = a^{prk_1+qsk_2}$. Since $\langle a \rangle = \langle a^{pr}a^{qs} \rangle$, it follow that $x\in\langle a \rangle$. Therefore $x\in\langle a^{pr}a^{qs} \rangle$.

Now suppose $x \in \langle a^{pr}a^{qs} \rangle$, then $x = (a^{pr}a^{qs})^k$ for some integer $k$. Thus, $x = (a^{pr})^k(a^{qs})^k$. Therefore $x \in \langle a^{pr} \rangle \langle a^{qs} \rangle$.

Since $\langle a \rangle$ has order $rs$, the number of generators of $\langle a \rangle = G$ is $\phi(rs)$.

By the other hand, since $\langle a^{pr} \rangle$ has orden $s$ and $\langle a^{qr} \rangle$ has orden $r$, the number of generators of $\langle a^{pr} \rangle \langle a^{qr} \rangle = G$ is $\phi(r)\phi(s)$.

Thus, $$\phi(rs)=\phi(r)\phi(s)$$

Related articles have already been read.

An Introduction to Theory of Groups by Joseph J. Rotman - Exercise 2.21 - (ii)

Question about proving $\phi(mn) = \phi(m)\phi(n)$

Answer 1

An alternative way to approach part (ii) is by noticing which elements of the form $b^j c^k$ with $1\leq j\leq r $ and $1\leq k\leq s $ are in $\text{gen}(G)$:

$$ |b^jc^k|=rs\Leftrightarrow |b^j||c^k|=rs$$ This last step follows from the fact that $gcd(|b^j|,|c^k|)=1$ and we have an abelian group.

$$|b^j||c^k|=rs\Leftrightarrow |b^j|=r \:\text{ and }\: |c^k|=s$$

$|b^j|=r\Leftrightarrow gcd(r,j)=1$. There are $\phi(r)$ values of $j$ such that this holds. Similarly, there are $\phi(s)$ values of $k$ such that this holds. Thus, there are at least $\phi(s)\phi(r)$ elements of order $rs$. We are almost done, but we still need to answer two questions:

Are we counting things twice?

Let $1\leq j\leq r $ and $1\leq k\leq s $. We have $ b^j c^k =a^t=b^t c^t=b^{rq_1+r_1}c^{sq_2+r_2}=b^{r_1}c^{r_2}$. We have $a^{x_os(j-r_1)}=b^{j-r_1}=c^{r_2-k}=a^{y_or(r_2-k)}$ (where we have writen $b=a^{x_os}$ ans $c=a^{y_or}$).Thus $rs|x_os(j-r_1)-ry_o(r_2-k)$ and we have $s|sx_o(j-r_1)+ry_o(k-r_2)\Rightarrow s|(k-r_2)$ and $c^{r_2-k}=1$. We have $1\leq r_2, k\leq s $, so the only way for this to happen is that $r_2=k$. Similarly $r_1=j$.

Thus, we are not counting things twice!

Are all generator elements there?

If $|a^j|=rs$, we clearly have $a^j=b^j c^j=b^{r_{1}}c^{r_2}$, where $r_1$ and $r_2$ are the remainders when divided by r and s respectively so it must have been counted.