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If $H,K$ are subgroups of $G$ s.t. $o(H), o(K)$ are relativily prime $\implies H \cap K = \{ e \}$.

Here $o(H)$ means the order of $H$.

Below is my attempt but I am afraid I might be jumping some hoops here:

Suppose $a \in H \cap K$ is an arbitrary element. Then we know $o(a)$ must divide $o(H)$ and $o(K)$.

Then $\exists m,n \mathbb{N}$ s.t. $m \cdot o(a) = o(H)$ and $n \cdot o(a) = o(K)$.

But $o(H)$ and $o(K)$ are relatively prime, so for $m \cdot o(a)$ and $n \cdot o(a)$ to be reelatively prime, $o(a)$ must be equal to $1$ which means $a=e$ the identity element.

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  1. Am I on the right track?
  2. Now my problem is with the last statement. How do I effectively argue the last statement?
  3. Any alternative proof?
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    $\begingroup$ Your proof is perfectly fine. You need not introduce the numbers $m$ and $n$ to write down the proof actually : two integers are relatively prime if and only if their only common divisor is $1$ (well, and $-1$). As soon as you know that $o(a)$ divides both $o(H)$ and $o(K)$, you can conclude $o(a)=1$. $\endgroup$ – Suzet Feb 5 at 5:10
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    $\begingroup$ You solution is correct, however, I am giving a more compact solution below, although, more or less these two solutions are the same. $\endgroup$ – aud098 Feb 5 at 5:25
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Note, $H\cap K$ is group and moreover subgroup(since, intersection of two subgroup of a same group is again a group ) of both $H, K$. So, $o(H\cap K)|o(H), o(H\cap K)|o(K)$, but since $o(K), o(H)$ are co-primes, so $1$ is the only common divisor to them.

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By Lagrange's theorem, the order of $H \cap K$ must divide the orders of both $H$ and $K$. Since the orders of $H$ and $K$ are relatively prime, this means that $H \cap K$ must be trivial.

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