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if $G$ is a finite group. $H,K$ are subgroups of $G$ and $|H|=18$, $|K|=25$ than why the intersection of $H$ and $K$ is only the unit element and can't include more elements?

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What do you know about the intersection of two subgroups? What do you know about the order of a subgroup? – Gerry Myerson Jan 16 '13 at 11:01

The intersection of any two subgroups of $G$ is again a subgroup of $G$. By lagrange's theorem $|H\cap K|| |H|,|K|$ . What is $\gcd(|H|,|K)$

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I guess it something with Lagrange but why if the gcd=1 we get this? – Anna Jan 16 '13 at 11:03
oh thank you now I understand:)! – Anna Jan 16 '13 at 11:05
$|H\cap K|$ is common divisor of $18,25$. The only common divisor of $18,25$ is $1$ – Amr Jan 16 '13 at 11:05

If $G$ is a group and $H,K$ be two subgroups of it, as @Amr noted, you can prove that $H\cap K$ is a subgroup of $G$ containing in both $H$ and $K$. So $$H\cap K\subseteq H~~\text{and}~~H\cap K\subseteq K$$ so according to Lagrange theorem since $H$ and $K$ are both finite so the order of $H\cap K$,say $t$, is finite and divides $|H|$ and $|K|$. Now find the number that divides $25$ and $18$ simultaneously. It is just $t=1$.

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Very neat and straightforward! +1 – amWhy Feb 15 '13 at 0:05

Hint: Suppose $x\in H\cap K$. Use Lagrange's theorem for $\langle x\rangle$.

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I think Lagrange is overkill. – Gerry Myerson Jan 16 '13 at 11:02
Lagrange is the most fundamental theorem in the theory of finite groups. How could this possibly be an overkill? – Curufin Jan 16 '13 at 11:10
I don't think Lagrange is overkill after the first 2-3 hours of group theory studies: in most books it is introduced in the first 1-2 chapters and the OP's comments below other question show she's acquainted with it. – DonAntonio Jan 16 '13 at 11:19
@Curufin, my apologies --- I was misunderstanding Lagrange's Theorem to be the one about the order of an element dividing the order of the group, whereas I now realize it's the one about the order of a subgroup dividing the order of the group. – Gerry Myerson Jan 17 '13 at 3:11

Since you did not choose any answer yet, here is more "physical" approach. The purpose of writing this is not to have an efficient answer.

If we have a nontrivial $x \in H \cap K$, then $\langle x \rangle$ has order $5$ or $25$ by Lagrange. This contradicts the fact that this cyclic subgroup is contained in $H$.

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