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

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$$ $$ 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.

Answer 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.

1. $x \neq y$
2. $x = y$

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.

Answer 2

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...

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