Gaps in the proof finite subgroup of the multiplicative group of a field must be cyclic

Question

I am trying to prove that a finite subgroup of the multiplicative group of a field must be cyclic. I got a proof in one of the textbook for Algebra(attached below). But I cannot see how the last step jumps from $m=lcm(d_1,d_2,...,d_r)=d_1d_2...d_r$ to the fact that $G$ is isomorphic to $\Bbb Z_m$.

I have some ideas but it is not complete and may be wrong:

Since $lcm(d_1,d_2,...,d_r)=d_1d_2...d_r$, it may be true(true?why?) that $gcd(d_i,dj_)=1$ for all $i\neq j$, thus $\Bbb Z_{d_1}\times...\Bbb Z_{d_r}$ is cyclic and $G\cong\Bbb Z_{d_1}\times...\Bbb Z_{d_r}$ is cyclic.

Is my idea correct? if so, can anyone help fill the gaps?

Answer 1

Your idea is correct. At the beginning of the proof is stated that $G\cong\Bbb Z_{d_1}\times...\Bbb Z_{d_r}$, where $d_i$ is a power of a prime for every $i$. Because of this, $G$ is cyclic if and only if $(d_i,d_j)=1$ for every $i \neq j$ (You can prove this as an exercise). Now consider that if for some distinct $i,j$ the numbers $d_i$ and $d_j$ share a prime factor, then $m=lcm( d_1, d_2, \dots d_r)$ is strictly less than $ d_1 d_2 \dots d_r$. Therefore, if these two numbers are equal, $(d_i,d_j)=1$ for every $i \neq j$ .

Answer 2

My answer:

if $gcd(d_i,d_j)>1$ for some $i\neq j$, then they has a common prime factor p. Thus, $\frac {d_1d_2...d_r}{p}$ is an integer, hence, a common multiplier of $d_1,...,d_r$, which implies $lcm(d_1,...,d_r)\le\frac {d_1d_2...d_r}{p}<d_1d_2...d_r$. Thus $gcd(d_i,d_j)=1\forall i\ne j$.

Then, the order of $(1,...,1)$ in $\Bbb Z_{d_1}\times...Z_{d_r}$ is $lcm(d_1,...,d_r)=d_1...d_r=|Z_{d_1}\times...Z_{d_r}|$, therefore, $G≅Z_{d_1}\times...Z_{d_r}$ is cyclic