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

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.

$$\Box$$

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?
• 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 $o(a)=1$. – Suzet Feb 5 at 5:10
• You solution is correct, however, I am giving a more compact solution below, although, more or less these two solutions are the same. – aud098 Feb 5 at 5:25

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.
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.