Answer to Question #139615 in Abstract Algebra for J

Question

if (H,*) is a subgroup of (G,*) and if R is the relation defined on G by aRb iff a * b^-1 is an element of H, then R is an equivalence relation on G, and that aRb iff b is an element of Ha (where Ha={h * a: h is an element of H}), and thus [a] subscript of R=Ha.Let G=S subscript of 4 , and let H=<{(13),(14)}>. Find H, and find all of the equivalence classes of S subscript of 4 under R.

Answer 1

Let $(H,*)$ be a subgroup of $(G,*)$ and if $R$ be the relation defined on $G$ by $aRb$ iff $ab^{-1}\in H$.

Show that $R$ is an equivalence relation on $G$.

Since $a*a^{-1}=e\in H$ for any $a\in G$, $R$ is reflexive relation.

If $aRb$, then $a*b^{-1}\in H$. Since $H$ is a subgroup, $b*a^{-1}=(a*b^{-1})^{-1}\in H$. Therefore, $bRa$ and $R$ is symmetric relation.

If $aRb$ and $bRc$ , then $a*b^{-1}\in H$ and $b*c^{-1}\in H$. Since $H$ is a subgroup, $a*c^{-1}=(a*b^{-1})*(b*c^{-1})\in H$ . Therefore, $aRc$ and $R$ is transitive relation.

Consequently, $R$ is an equivalence relation.

Taking into account that $b*a^{-1}\in H$ if and only if there exist $h\in H$ such that $b*a^{-1}=h$, i.e. $b=h*a\in Ha$, we conclude that $bRa$ iff $b$ is an element of $H*a=\{h*a|h\in H\}$. Therefore, $[a]_R=\{b\in G|bRa\}=H*a=\{h*a|h\in H\}$.

Consider subgroup $H=\langle{(13),(14)}\rangle=\{(1),(13),(14),(34),(134),(143)\}$ of the symmetric group $S_4$. Since $|S_4|=24$ , there exist $\frac{24}{6}=4$ different equivalence classes $S_4$ under $R$. Let us find them:

$[(1)]_R=H(1)=H=\{(1),(13),(14),(34),(134),(143)\},$

$[(12)]_R=H(12)=\{(1),(13),(14),(34),(134),(143)\}(12)=\{(12), (123), (124), (12)(34), (1234), (1243)\},$

$[(23)]_R=H(23)=\{(1),(13),(14),(34),(134),(143)\}(23)=\{(23), (132), (14)(23), (243), (1324), (1432)\},$

$[(24)]_R=H(24)=\{(1),(13),(14),(34),(134),(143)\}(24)=\{(24), (13)(24), (142), (234), (1342), (1423)\}.$