Understanding the Dicyclic Group of Order 12: Composition and Element Orders

[Azure Ace]

Homework Statement

The dicyclic group of order 12 is generated by 2 generators x and y such that: $y^2 = x^3, x^6 = e, y^{-1}xy =x^{-1}$ where the element of Dic 12 can be written in the form $x^{k}y^{l}, 0 \leq x < 6, y = 0,1$. Write the product between two group elements in the form $(x^{k}y^{l})(x^{m}y^{n})$ and show that $a^3$ is the only order 2 element and $a^2$ is the only order 3 element in Dic12.

Homework Equations

$y^2 = x^3, x^6 = e, y^{-1}xy =x^{-1}$

The Attempt at a Solution

The answer I was able to obtain is $(x^{k}y^{l})(x^{m}y^{n}) = x^{k-m}y^{l+n}$ for $l = -1, 1$ and $(x^{k}y^{l})(x^{m}y^{n}) = x^{k+m}y^{n}$ for $l = 0$. Is this correct? Also, Isn't it that $a^4$ also an order 3 element? So now, I'm confused T_T

 

[Dick]

Homework Statement

The dicyclic group of order 12 is generated by 2 generators x and y such that: $y^2 = x^3, x^6 = e, y^{-1}xy =x^{-1}$ where the element of Dic 12 can be written in the form $x^{k}y^{l}, 0 \leq x < 6, y = 0,1$. Write the product between two group elements in the form $(x^{k}y^{l})(x^{m}y^{n})$ and show that $a^3$ is the only order 2 element and $a^2$ is the only order 3 element in Dic12.

Homework Equations

$y^2 = x^3, x^6 = e, y^{-1}xy =x^{-1}$

The Attempt at a Solution

The answer I was able to obtain is $(x^{k}y^{l})(x^{m}y^{n}) = x^{k-m}y^{l+n}$ for $l = -1, 1$ and $(x^{k}y^{l})(x^{m}y^{n}) = x^{k+m}y^{n}$ for $l = 0$. Is this correct? Also, Isn't it that $a^4$ also an order 3 element? So now, I'm confused T_T

For one thing you seem to be using $x$ and $a$ to denote the same symbol. That's confusing. I do think your product relation looks like it works out. Though I'd choose to write it differently (I'm not sure what $l=-1$ is for). But sure, order 3 elements always come in pairs.

 

[Azure Ace]

For one thing you seem to be using $x$ and $a$ to denote the same symbol. That's confusing. I do think your product relation looks like it works out. Though I'd choose to write it differently (I'm not sure what $l=-1$ is for). But sure, order 3 elements always come in pairs.

Sorry, a is supposed to be $x$. I'm confused by the problem that $x^2$ is the only order 3 element, but I also think that $x^4$ is also an order 3. I thought that maybe my product is wrong or something.

 

[fresh_42]

The German Wikipedia page has a group table for $Dic_{12}$ which is called $Dic_3$ there.
https://de.wikipedia.org/wiki/Dizyklische_Gruppe#Dicn_für_kleine_n
I would prefer the composition: $Dic_{12} \cong \mathbb{Z}_6 \rtimes \mathbb{Z}_2 \cong (\mathbb{Z}_2 \times \mathbb{Z}_3) \rtimes \mathbb{Z}_2$.