Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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

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.

share|cite|improve this answer
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

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

share|cite|improve this answer
My GAPy friend. It is Easy. +1 – Babak S. May 4 '13 at 4:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.