MHB Find Left Cosets of Subgroup in $\mathbb{Z}_{15}, D_4$

[NoName3]

How do I find the left cosets of:

$(a)$ $\left\{ [0], [5], [10] \right\} \le \mathbb{Z}_{15}$ ($\mathbb{Z}_n$ is additive group modulo $n$).

$(b)$ $\left\{e, y, y^2, y^3 \right\} \le D_4$ where $y$ denotes rotation of a square.

The not equal to here denotes subgroup. The trouble I've with the first one is that I'm finding too many left cosets. The definition of left cosets is $\left\{gh: g \in G, h \in H\right\}$. In this case if I let $h = 0$ and vary $g$ through $0$ to $14$ then my set contains fifteen members. Well, that can't right surely? I think there's a theorem that says my subgroup can only have five left cosets (not too sure if that's right).

 

[Opalg]

How do I find the left cosets of:

$(a)$ $\left\{ [0], [5], [10] \right\} \le \mathbb{Z}_{15}$ ($\mathbb{Z}_n$ is additive group modulo $n$).

$(a)$ $\left\{e, y, y^2, y^3 \right\} \le D_4$ where $y$ denotes rotation of a square.

The not equal to here denotes subgroup. The trouble I've with the first one is that I'm finding too many left cosets. The definition of left cosets is $\left\{gh: g \in G, h \in H\right\}$. In this case if I let $h = 0$ and vary $g$ through $0$ to $14$ then my set contains fifteen members. Well, that can't right surely? I think there's a theorem that says my subgroup can only have five left cosets (not too sure if that's right).

In $(a)$, the group operation in $\mathbb{Z}_n$ is additive, so it is usual to use the notation $g+H$ rather than $gH$ for the cosets. If for example the group is $\mathbb{Z}_{15}$, the subgroup is $H = \left\{ [0], [5], [10] \right\}$ and we take $g$ to be the element $[1]$, then the coset $[1] + H$ consists of the elements $[1]+[0]$, $[1]+[5]$ and $[1]+[10].$ That gives you the coset $\left\{ [1], [6], [11] \right\}.$ If you carry on like that, you will find that there are indeed five cosets (each consisting of three elements).

 

[NoName3]

In $(a)$, the group operation in $\mathbb{Z}_n$ is additive, so it is usual to use the notation $g+H$ rather than $gH$ for the cosets. If for example the group is $\mathbb{Z}_{15}$, the subgroup is $H = \left\{ [0], [5], [10] \right\}$ and we take $g$ to be the element $[1]$, then the coset $[1] + H$ consists of the elements $[1]+[0]$, $[1]+[5]$ and $[1]+[10].$ That gives you the coset $\left\{ [1], [6], [11] \right\}.$ If you carry on like that, you will find that there are indeed five cosets (each consisting of three elements).

Thank you. I was counting the elements of the cosets, not cosets. I have now found that:

$g+H = \left\{ \left\{[0], [5], [10]\right\}, \left\{[1], [6], [11]\right\}, \left\{[2], [7], [12]\right\}, \left\{[3], [8], [13]\right\}, \left\{[4], [9], [14]\right\} \right\}.$

Regarding $(2)$ is it possible to define $D_4$ the way $D_6$ is defined in here i.e. $D_6 = \left\{x^i, yx^i: 0 \le i \le 5 \right\}$?

 

[Opalg]

Thank you. I was counting the elements of the cosets, not cosets. I have now found that:

$g+H = \left\{ \left\{[0], [5], [10]\right\}, \left\{[1], [6], [11]\right\}, \left\{[2], [7], [12]\right\}, \left\{[3], [8], [13]\right\}, \left\{[4], [9], [14]\right\} \right\}.$

Correct. (Yes)

Regarding $(2)$ is it possible to define $D_4$ the way $D_6$ is defined in here i.e. $D_6 = \left\{x^i, yx^i: 0 \le i \le 5 \right\}$?

Yes, except that there is a clash of notation. The $y$ in your $D_4$ corresponds to the $x$ in Wolfram's $D_6$. So your $D_4$ should consist of elements $e,\,y,\,y^2,\,y^3,\,z,\,zy,\,zy^2,\,zy^3$, satisfying the relation $yz = zy^3.$

If you are viewing the group as symmetries of a square then $y$ represents a rotation of $90^\circ$ and $z$ represents reflection in a line joining the midpoints of two opposite edges.

 

[NoName3]

Correct. (Yes)Yes, except that there is a clash of notation. The $y$ in your $D_4$ corresponds to the $x$ in Wolfram's $D_6$. So your $D_4$ should consist of elements $e,\,y,\,y^2,\,y^3,\,z,\,zy,\,zy^2,\,zy^3$, satisfying the relation $yz = zy^3.$

If you are viewing the group as symmetries of a square then $y$ represents a rotation of $90^\circ$ and $z$ represents reflection in a line joining the midpoints of two opposite edges.

Thanks again!

I think in finding $gH$ letting $g = e$ we find the first coset $\left\{e, y, y^2, y^3\right\}$ and letting $g = z$ we get $\left\{z, zy, zy^2, zy^3\right\} =\left\{z, y^3z, y^2z, yz\right\}. $

Lagrange's theorem says that the number of left cosets is exactly $|G|/|H|$ which is $8/4 = 2$ here so there are none more to find.