Let $G=\langle x, y\mid x^2, y^2\rangle$. Understanding proof $\exists H\unlhd G$ s.t. $[G:H]=2$.

Question

I met some problem during googling.

The problem and its solution are next.

and I'm wondering about 2nd YELLOW BOX

$$ $$

enter image description here

$$ $$ Why does $G$ have a unique element of order $2$ in case of $H=G$ ?

$$ $$

The next is "my opinion" for 2nd YELLOW BOX...

If $H=G=H \cup Hx$ , then $Hx=H$

So, $x=h^k$ for some $k \in Z-\{0\} $.

By hypothesis for $|x|=2$ , I get $|h^k|=2$.

Therefore $G=\{1, h, h^2, \cdots , h^k , ... , h^{2k-1} \}$, and $G$ has an unique element $h^k$ of order $2$.

Is my opinion RIGHT?

I thought and thought AND THOUGHT so many time for digesting YELLOW BOX.. :-(

By any chance... Is it trivial thing?

OR Do you have any good idea for understanding YELLOW BOX?

Could you give me some advice, please?

Thanks for reading my question.

The first yellow box is just the definition of $H=\langle h\rangle$. — lhf, Oct 21 '14 at 10:32
Oops.. I'm just wondering 2nd YELLOW BOX.. Sorry,, why I made two boxes...sorry — user143993, Oct 21 '14 at 12:33

score 5 · Answer 1 · edited Apr 13 '17 at 12:19

5

You said that $Hx\subset H$, but remember cosets are disjoint or equal! So $Hx=H$. Then, remembering $H=\langle h\rangle$ where $h=xy$, and assuming $Hx=H$,

$y\in H$ as $h^{-1}x\in Hx=H$ but $h^{-1}x=y^{-1}$.
$x\in H$ as $1x\in Hx=H$ but $1x=x$.

Hence, $G=\langle x, y\rangle\leq H$, so $G=H$ as required.

As $H$ is cyclic, this means that $G$ is cyclic. Now, cyclic groups of order $n$ contain unique subgroups of every order dividing $n$$^{\dagger}$, from which the "unique element of order $2$" follows. Alternatively, note that the number $n$ has at most one integer that is one half of it...

$^{\dagger}$ See this math.SE answer for a proof of this fact.

edited Apr 13 '17 at 12:19

Community

1

answered Oct 21 '14 at 10:16

user1729

31,015

And I edited Hx=H, thanks. – user143993 Oct 21 '14 at 12:40
Thanks for answering. BUT I'd like to know what make you said 'the order of G is n(i.e. finite)' actually I wonder 2nd YELLOW BOX... Can 'my opinion(the gray box)' explain the finite order of G? – user143993 Oct 21 '14 at 12:50
@user I only assumed that $G$ had order $n$ when I had proven that it was cyclic. As $\mathbb{Z}$ contains no elements of order 2, $G$ must be finite cyclic. – user1729 Oct 21 '14 at 15:00
Aha thank you very much. Have a nice day. – user143993 Oct 22 '14 at 02:22

Simon Rose · Accepted Answer · 2014-10-21T10:58:39.507

1

There's a quick answer to this: If $H = G$, then since $H$ is cyclic (it is generated by the element $h = xy$), $G$ must also be cyclic. Since it is cyclic, it is in particular abelian. Thus $x, y$ commute.

So we have two possibilities.

$x \neq y$
$x = y$

You should be able to show that the first case is impossible since $G$ is cyclic. This leaves the second case, but this shows that there is only one element of order 2.

Edit This is not correct, as was pointed out in the comments. A better proof is that if $G$ is cyclic (by hypothesis, since $G = H$), then $G \cong \mathbb{Z}$ or $G \cong \mathbb{Z}/n$ for some $n$. In the first case, it has no elements of order 2, which obviously is not the case. In the second case, if $n$ is odd then there are also no elements of order 2. If $n$ is even, then you can check (try this with $\mathbb{Z}/6, \mathbb{Z}/8$ and a few others if you prefer) that there must be a unique element of order two, namely $[n/2]$. Since by assumption $x \neq y$, here is your contradiction.

edited Oct 21 '14 at 10:58

answered Oct 21 '14 at 10:36

Simon Rose

6,793

Yeah, fair enough. I (admittedly) didn't read the whole question, but just the highlighted bit. Ooops! – Simon Rose Oct 21 '14 at 10:59
Thanks for your answering. Could you read my opinion [the gray box]? Is that right thing? If I write right, then in case H=G.. the order of G must be finite, isn't it? Thank you. – user143993 Oct 21 '14 at 12:42
and I focused on 'Hx=H". Thanks. – user143993 Oct 21 '14 at 12:46
Yes, what you have written in the grey box is correct. Essentially, it is the same sort of argument that I have written in my edit. – Simon Rose Oct 21 '14 at 12:53
Aha, in the now I understand certainly what you said. THANKS – user143993 Oct 21 '14 at 13:10

Let $G=\langle x, y\mid x^2, y^2\rangle$. Understanding proof $\exists H\unlhd G$ s.t. $[G:H]=2$.

2 Answers2