Question about grp of order 6...

[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?

[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?

[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.

[micromass]

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.

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??

[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.

[micromass]

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.

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.

[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?

[micromass]

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?

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

[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.

[micromass]

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.

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

[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'.

[micromass]

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'.

Yes!

[Samuelb88]

Thank you very much for all your help, micromass!