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So I supposed $|H \cap K|>1$

$\Rightarrow |HK||H \cap K|> |HK|$

Which eventually implied that

$\Rightarrow |H \cap K|>|G|$

Thus since G is a group, and H and K are subgroups then the identity belongs to both H and K. Since it belongs to both H and K then it belongs to $H \cap K$. But if $|H \cap K|>|G|$ then that implies $|H \cap K|$ is greater than one meaning it has at least two elements.

Im not sure if thats the outcome I should get but I was wondering if there are other ways to prove this problem.

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If you want to prove $|H\cap K|\geq 2$, why do you begin with, "Suppose $|H\cap K|>1$?" –  Clayton May 4 '13 at 2:02

2 Answers 2

Sanity check: How can $\;|H \cap K|>|G|\;?\;$...if both $H \leq G$ and $K\leq G$.

It is never a good practice to prove that $X$ by assuming that $X$. You began with

"Suppose $|H \cap K|>1."\;$ In doing so, you are assuming precisely what you need to prove!

One approach: To prove that $|H\cap K| \geq 2$, assume for the sake of contradiction that $|H\cap K| = 1$: that there intersection contains only the identity element in $G$.

You'll also want to make use of the premises: $\;|H|^{2}>|G|,\; |K|^2 > |G|.\;$ We want to see what follows from these facts, given $H\leq G, K\leq G.\;$ Your intuition about using $|HK|$ in the proof is spot on. If we can conclude that the intersection of $H$ and $K$ must be nontrivial, we are done.

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Let me know if you have further questions...I'll help you work it out... (-: –  amWhy May 4 '13 at 2:58
    
>8-) But I cannot make additional plus, Amy..... –  Babak S. May 4 '13 at 4:22
    
Hello, Babak! Just in time, I see you, before I go to bed! 8D –  amWhy May 4 '13 at 4:25
    
Indeed, it makes me strength of attacking problems. I mean, greeting you before diving into warm bed. :-) –  Babak S. May 4 '13 at 4:28

$$|H\cap K|^2=\frac{|H|^2|K|^2}{|HK|^2}>\frac{|G|^2}{|G|^2}=1\Rightarrow|H\cap K|>1$$

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My GAPy friend. It is Easy. +1 –  Babak S. May 4 '13 at 4:24

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