Let $G$ a finite group and $H\leq G$, such that $|H| = \frac{|G|}{2}$. If $x \in G$ then $x^2 \in H$

Asked Mar 17 '22 at 19:44

Active Apr 05 '22 at 06:20

Viewed 62 times

My attempt:

As $x \in G$ then $\operatorname{ord}(x)$ divide $|G|$ this implies that $|G| = k\cdot \operatorname{ord}(x)$ with $k\in \mathbb{N}$

Moreover, we know that

$$\operatorname{ord}(x^2)=\frac{\operatorname{ord}(x)}{\gcd(2,\operatorname{ord}(g))}$$

Then:

$$\operatorname{ord}(x^2)=\frac{|G|}{k\cdot \gcd(2,\operatorname{ord}(g))}$$

Here I'm stuck. Can someone help me?

edited Apr 05 '22 at 06:20

asked Mar 17 '22 at 19:44

rcoder

2

See also this duplicate. The hint was: If $x^2H = xH$, then $xH = H$. – Dietrich Burde Mar 17 '22 at 19:48
Hello, thanks for your answer. But i don't see why $x^2H = xH$ can you give me a hint of this? @DietrichBurde – rcoder Mar 17 '22 at 20:33
1

Use that the index is $2$, i.e, we have only two cosets. – Dietrich Burde Mar 17 '22 at 20:40
ok thanks!!!!@DietrichBurde – rcoder Mar 17 '22 at 21:10
If $[G:H]=2$, than $H\unlhd G$ and $G/H\cong C_2$. Therefore, for $a\in G\setminus H$, $aH$ has order $2$ in the quotient, i.e. $(aH)^2=a^2H=H$, whence $a^2\in H$. If $a\in H$, $a^2\in H$ by closure. Therefore, $a^2\in H$ for every $a\in G$. – Apr 03 '22 at 16:09

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