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