# Explain why D3 cannot be a subgroup of D8

To be a subgroup, a subset of a group must satisfy the group axiom but in this case, I do not see how the group axiom plays a part.

Could someone explain to me why the above question is true?

• What is your definition of $D_3$ and $D_8$, explicitly? – Zev Chonoles Jun 26 '15 at 3:13
• Dihedral group of degree 3 and degree 8 – Mathematicing Jun 26 '15 at 3:18
• Right, I know that - I mean, there are multiple possible definitions of "dihedral group" (all ultimately equivalent, of course). Are you thinking of them in terms of rotations, group presentations, etc.? – Zev Chonoles Jun 26 '15 at 3:37

The explanation is immediate from Lagrange's theorem. If you haven't seen that yet, are you sure that $D_3 \subset D_8$?

• I am not sure, that's why I was hoping for clarifications. – Mathematicing Jun 26 '15 at 3:20
• To help out a little bit: en.wikipedia.org/wiki/Lagrange%27s_theorem_%28group_theory%29 Basically the issue is that $16/6$ is not an integer. The left cosets of the subgroup have to "partition" the space evenly. – muaddib Jun 26 '15 at 3:25
• Ok, well, $D_3$ contains an element of order $3$. Does $D_8$? If you're not sure try writing out the group structure or playing with the symmetries of an octagon. – GiantTortoise1729 Jun 26 '15 at 3:26

By Lagrange's theorem we know that order of subgroup must divides order of group but here D_3 has order 6 while D_8 has order 16 and clearly 6 does not divides 16. so D_3 cant be subgroup of D_8

First, a group is a set together with a binary operation that satisfy some given conditions. On the other hand, $H$ is a subgroup of a group $G$ if $H$ is a subset of $G$ and satisfy all the properties of a group, under the same binary operation as $G$.

$D_3:=\langle x,y \mid x^3=y^2=1, xy=yx^{-1} \rangle = \{1,x,x^2,y,xy,x^2y\}$ while $D_8:=\langle x,y \mid x^8=y^2=1, xy=yx^{-1} \rangle = \{1,x,x^2,x^3,x^4,x^5,x^6,x^7,y,xy,x^2y,x^3y,x^4y,x^5y,x^6y,x^7y\}.$

For instance $x^2$ has order $3$ in $D_3$ while same $x^2$ has order $4$ in $D_8$. The reason why $x^2$ does not have the same order in both $D_3$ and $D_8$ is that the binary relation in both $D_3$ and $D_8$ are not the same. This is a first reason why $D_3$ is not a subgroup of $D_8$. In other words, $D_3$ is not a subgroup of $D_8$ because $D_3$ is not a subset of $D_8$ under the same binary operation.

First, a group is a set together with a binary operation that satisfy some given conditions. On the other hand, $H$ is a subgroup of a group $G$ if $H$ is a subset of $G$ and satisfy all the properties of a group, under the same binary operation as $G$.

$D_3:=\langle x,y \mid x^3=y^2=1, xy=yx^{-1} \rangle = \{1,x,x^2,y,xy,x^2y\}$ while $D_8:=\langle x,y \mid x^8=y^2=1, xy=yx^{-1} \rangle = \{1,x,x^2,x^3,x^4,x^5,x^6,x^7,y,xy,x^2y,x^3y,x^4y,x^5y,x^6y,x^7y\}.$

$D_3$ is not a subgroup of $D_8$ because $D_3$ is not a subset of $D_8$ under the same binary operation.

For instance $x^2$ has order $3$ in $D_3$ while $x^2$ has order $4$ in $D_8$. The reason why $x^2$ does not have the same order in both $D_3$ and $D_8$ is that the binary relation in both $D_3$ and $D_8$ are not the same.