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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?

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  • $\begingroup$ 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
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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 $$I \cap J = (I \cap J)(I + J) = (I \cap J)I + (I \cap J)J \subseteq JI+IJ=IJ.$$

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