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. See here if you are not familiar with this result.

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

$|b^j|=r\Leftrightarrow gcd(r,j)=1$. 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.

Answer 2

The approach I imagine the question intended was as follows:

Suppose that $G$ is a cyclic group of order $n=r.s$ where $\text{g.c.d.}(r,s)=1$ and let $a$ be a generator of $G$, so that $o(a)$, the order of $a$, is $n$.

Claim 1: If $g=a^k \in G$ for some $k \in \{0,1,\ldots,n-1\}$ then $o(g) = \frac{n}{\text{g.c.d.}(n,k)}$. In particular, $G$ has $\phi(n/d)$ elements of order $d$ for each $d \mid n$.

Proof: If $d=o(g)$ then $g^m=1$ if and only if $d\mid m$. But $$ g^m = a^{k.m}=1 \iff n\mid k.m \iff \frac{n}{\text{g.c.d.}(k,n)} \mid m, $$ and hence it follows that $o(a^k) = \frac{n}{\text{g.c.d.}(k,n)}$. For the final sentence, note that if $a$ has order $n$ then the elements of order $d$ for $d\mid n$ are those of the form $a^{(n/d)t}$ where $t \in \{k \in \{0,1,\ldots,d-1\}: \text{g.c.d.}(k,d)=1\}$.

Claim 2 If $g \in G$ has order $n$, then there are unique $h,k \in G$ with $g=hk$ and $o(h)=r,o(k)=s$. Moreover $g\mapsto (h,k)$ gives a bijective correspondence between $$ \{g \in G: o(g)=n\} \longleftrightarrow \{(h,k)\in G\times G: o(h)=r, o(k)=s\} $$

Proof: Since $\text{g.c.d.}(r,s)=1$, there are integers $p,q$ such that $pr+qs=1$. It follows that if $g$ has order $n$ then setting $h=g^{qs}, k = g^{pr}$ it is easy to see that $h^r =k^s = 1$ and $hk=g$. On the other hand, if $g = hk$ where $h$ and $k$ have orders $r$ and $s$ respectively, then $$ g^{qs}=(hk)^{qs} = h^{1-pr}k^{qs} = h(h^r)^{-p}.(k^s)^q =h. $$ Similarly one finds that $k=g^{pr}$ and so the pair $(h,k)$ is unique. For the bijectivity of $g \mapsto (h,k)$ note that in any group $H$ containing elements $h_1,h_2$ which commute, that is, $h_1h_2 = h_2 h_1$ the order $o(h_1h_2)$ of $h_1h_2$ is the least common multiple of $o(h_1)$ and $o(h_2)$.

Now part $ii)$ follows immediately: The bijective correspondence of Claim 2 shows that the cardinalities of the two sets are equal, and Claim 1 shows that these are $\phi(n)$ and $\phi(r)\phi(s)$ respectively.