Question about grp of order 6

1. Jul 18, 2011

Samuelb88

Lemma. A group $G$ of order 6 can have only one element of order 3.

Pf. Suppose $G$ has two elements of order 3. Call these elements $x$ and $y$. Let $H$ and $K$ be the subgroups generated by $x$ and $y$ resp. Then $H \cap K = \{ e \}$ and therefore $G$ can have only one subgroup of order 3.

I'm reading over my notes from class and I'm confused on the reasoning here. Why does $H \cap K = \{ e \}$ imply that $G$ can have only one element of order 3?

2. Jul 18, 2011

micromass

How many elements does

$$HK=\{hk~\vert~h\in H,k\in K\}$$

have if $H\cap K=\{e\}$. That is, can you find more than 6 elements in this group?

3. Jul 18, 2011

Samuelb88

Hi micromass.

Let $H = \{ e, x, x^2 \}$, and let $K = \{ e, y, y^2 \}$. If $H \cap K = \{ e \}$, then $HK = \{ e, xy, x^2y, x^2y, x^2y^2, yx, yx^2, y^2x, y^2x^2 \}$ and $|HK| = 9$. And since $e, x, x^2, y, y^2$ are unique, therefore there products constitute unique elements. Thus $G$ can have only one element of order 3.

4. Jul 18, 2011

micromass

No, that's not correct. Why can't xy=yx, for example?? Furthermore, you forgot x,y,x2 and y2.

Let's consider this set:

$$\{e,x,x^2,y,xy,x^2y,y^2xy^2x^2,y^2\}$$

Can you show that all of these elements are distinct from each other??

5. Jul 18, 2011

Samuelb88

Bummer. That was my wishful reasoning there. Well by cancellation laws, I can determine that:

1. $x \neq x^2$ since if they were equal, this would imply $x=e$, but $x$ has order 3.
2. $x \neq y$, by assumption.
3. $x \neq xy$ since $y \neq e$.
4. $x \neq x^2 y$ since if they were equal, by cancellation laws, it would follow that $x^{-1} = x^2 = y$, which can't happen since $x$ and $y$ generate unique elements.
5. $x \neq y^2 x y^2 x^2$ since if they were equal, it would follow that $y=e$.
6. $x \neq y^2$ since $x$ and $y$ generate unique elements.

A argument that the other elements are not equal is similar.

6. Jul 18, 2011

micromass

Uuuh, ok, that's good. But you can do things a lot easier. We just need to prove that

$$hk\neq h^\prime k^\prime$$

for h,h' in H and k,k' in K. But if equality holds then

$$h^\prime h =k^\prime k\in H\cap K$$

that's what your professor wants you to do.

7. Jul 18, 2011

Samuelb88

Okay, I think I see how to do this. Suppose $H \cap K = \{ e \}$. I want to suppose that there exists $h, h' \in H$ and $k, k' \in K$ such that $hk = h'k'$. It follows that $h^{-1}h' = kk'^{-1}$. Let $h'' = h^{-1}$ and $k'' = k'^{-1}$ and get $h'' h' = k k''$. But this would imply both $h'' h$ and $k k''$ belong to $H \cap K$, a contradiction. How does this look?

8. Jul 18, 2011

micromass

Why is this a contradiction? Why can't both h''h and k'' be in $H\cap K$?

9. Jul 18, 2011

Samuelb88

I was thinking that it was contradiction to my assumption that $H \cap K = \{ e \}$, but now I realize that $h^{-1} h'$ could equal the identity. But it seems strange to me that for different elements $h, h', k, k'$ that if $hk = h'k'$ then $h h'^{-1}$ could be the identity because wouldn't that be saying h=h' and k=k'? That is why I thought that if I showed that $hk=h'k'$ implied $h'' h'$ and $kk''$ belonged to $H \cap K$, it would contradict my supposition since I was assuming $h'' h' \neq e$.

Last edited: Jul 18, 2011
10. Jul 18, 2011

micromass

Yes, indeed, so if hk=h'k', then you have h=h' and k=k'. Wasn't this what you needed to show?

11. Jul 18, 2011

Samuelb88

So what you're saying is instead of citing a contradiction, what I should of concluded from my proof above that since $h'h''$ and $kk''$ belong to $H \cap K$, it follows that $h'h'' = e = kk''$ since $H \cap K = \{ e \}$. Thus $h=h'$ and $k=k'$.

12. Jul 18, 2011

micromass

Yes!

13. Jul 18, 2011

Samuelb88

Thank you very much for all your help, micromass!