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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 $\phi(mn) = \phi(m)\phi(n)$

Shaun
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Blue Tomato
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    $b$ generates the cyclic group $C_r$, and $a$ generates $C_s$, and $|U(C_r\times C_s)|=\phi(r)\phi(s)$, whereas $ |U(C_{rs})|=\phi(rs)$. This is just the same as the last link, and seems to be the easiest way. – Dietrich Burde Mar 11 '23 at 11:09
  • @DietrichBurde after a while, I confirmed that the proof exhibited is correct. The fine detail of the proof is that no type of isomophism is established with specific groups, that is, the proof works in general for groups that satisfy the hypotheses. – Blue Tomato Mar 13 '23 at 17:28

1 Answers1

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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.

Kadmos
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