If $H$ and $K$ are subgroups of finite index of a group $G$ such that $[G:H]$ and $[G:K]$ are relatively prime, then $G=HK$.

We know that $[G: H \cap K ] \le [G : H] \cdot [G : K]$, and that both $[G : H]$ and $[G : K]$ divide $[G: H \cap K ]$, implying that the least common multiple is no bigger than $[G: H \cap K ]$. Now, since the product of two numbers is equal to the product of the GCD and LCM, we have

$$[G : H] \cdot [G : K] = gcd([G : H], [G : K]) \cdot lcm([G : H] \cdot [G : K]$$

$$= lcm([G : H], [G : K] \le [G : H \cap K]$$

Thus, we have $[G: H \cap K ] \le [G : H] \cdot [G : K]$ and $[G : H] \cdot [G : K] \le [G : H \cap K]$. This implies they are equal which happens if and only if $G = HK$.

Does this seem right?

  • $\begingroup$ You know the least common multiple is $[G:H][G:K]$, so there's no need to go though the product of the lcm and gcd, you can get what you need from the second line. $\endgroup$ – Michael Burr Apr 17 '17 at 17:16
  • $\begingroup$ @MichaelBurr According to you, what is the second line? Prima facie, I don't believe I know $[G:H][G:K]$ is $lcm([G:H],[G:K])$ until I write the line "$[G : H] \cdot [G : K] = gcd([G : H], [G : K]) \cdot lcm([G : H] \cdot [G : K]$" and note that the GCD is $1$. $\endgroup$ – user193319 Apr 17 '17 at 17:29
  • $\begingroup$ Duplicate of math.stackexchange.com/questions/372979/… $\endgroup$ – Suman Kundu Apr 17 '17 at 18:39

Your proof is fine, but you don't have to use the equality $n \cdot m = \gcd(n,m) \cdot \mathrm{lcm}(n,m)$ (this is what Michael Burr hinted at). If $n,m$ divide $d$ and $n,m$ are coprime, then $n \cdot m$ also divides $d$. A quick proof is either prime factorization, or using the fact that coprime ideals $I,J$ of a commutative ring satisfy $I \cdot J = I \cap J$. This is because $$I \cap J = (I \cap J)(I + J) = (I \cap J)I + (I \cap J)J \subseteq JI+IJ=IJ.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.