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Question:

Let $H$ be a group, and $x, y, z \in H$.

  1. Prove that $\mathrm{ord}(xyx^{-1}) = \mathrm{ord}(y)$.
  2. Prove that $\mathrm{ord}(xy) = \mathrm{ord}(yx)$
  3. Assume that $H$ is also abelian, show that $Q_m = \{x \in H: \mathrm{ord}(x) \mid m\}$ is a subgroup of $H$ for any positive integer $m$.

My attempts are as follows:

For part (1), I am stuck, because from logic, I know that the operations done after $x$, $y$ and then inverse of $x$ (namely $x^{-1}$), is essentially doing the operation of $y$ alone, which is the right hand side of the statement. But I don't know how to write that out explicitly.

Or is there any formulas for ordering, which was not included in my lecture notes, will there be something like, i.e., $\mathrm{ord}(ab) = \mathrm{ord}(a)\mathrm{ord}(b) = \mathrm{ord}(b)\mathrm{ord}(a)$?

For part (2), as $H$ is not an abelian group in general, which means $xy$ not equal $yx$ in general, but from logic, I know that the $\mathrm{ord}(xy) = \mathrm{ord}(yx)$.

Because assume that $\mathrm{ord}(x) = 2$ and $\mathrm{ord}(y) = 3$, then $\mathrm{ord}(xy) = 6$, and also $\mathrm{ord}(yx) = 6$ as well, so $\mathrm{ord}(xy)=\mathrm{ord}(yx)$, but I don't know how to write that out explicitly.

For part (3), to show $Q_m$ is subgroup of $H$, so I need to check on $2$ things, closure and inverse.

For check of closure, I need to show that for all $x, y \in Q_m$, $x\cdot y \in Q_m$, but I have no clue of what operation, i.e. $\cdot $ is for group $H$ or group $Q_m$.

For check of inverse, I need to show that for all $x \in Q_m$ , the inverse of $x$ is also in $Q_m$, but I have no clue how to find the inverse of $x$ as well, or simply state it exists?

Please give me a little help, thank you!

Shaun
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Ramesh Karl
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1 Answers1

2

See my comments above for 1 and 2.

As for 3, I will use the one-step subgroup test.

Fix $m\in\Bbb N$.

Since $e\in H$, ${\rm ord}(e)=1$ and $1\mid m$, we have $e\in Q_m$. Thus $Q_m$ is nonempty.

By definition, $Q_m=\{\color{red}{x\in H}\; :\; {\rm ord}(x)\mid m\},$ so we have $Q_m\subseteq H$.

Let $x,y\in Q_m$. Then ${\rm ord}(x), {\rm ord}(y)\mid m$. Since the order of the inverse of an element is the order of the element, we have

$${\rm ord}(xy^{-1})\mid \frac{{\rm ord}(x){\rm ord}(y)}{{\rm gcd}({\rm ord}(x),{\rm ord}(y))}={\rm lcm}({\rm ord}(x),{\rm ord}(y))\mid m,$$

since $H$ is abelian. But $xy^{-1}\in H$ as $H$ is a group. Hence $xy^{-1}\in Q_m$.

Hence $Q_m\le H$.

Shaun
  • 44,997