Question #139615
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.
1
Expert's answer
2020-10-25T18:34:46-0400

Let (H,)(H,*) be a subgroup of (G,)(G,*) and if RR be the relation defined on GG by aRbaRb iff ab1Hab^{-1}\in H.


Show that RR is an equivalence relation on GG.

Since aa1=eHa*a^{-1}=e\in H for any aGa\in G, RR is reflexive relation.

If aRbaRb, then ab1Ha*b^{-1}\in H. Since HH is a subgroup, ba1=(ab1)1Hb*a^{-1}=(a*b^{-1})^{-1}\in H. Therefore, bRabRa and RR is symmetric relation.

If aRbaRb and bRcbRc , then ab1Ha*b^{-1}\in H and bc1Hb*c^{-1}\in H. Since HH is a subgroup, ac1=(ab1)(bc1)Ha*c^{-1}=(a*b^{-1})*(b*c^{-1})\in H . Therefore, aRcaRc and RR is transitive relation.

Consequently, RR is an equivalence relation.


Taking into account that ba1Hb*a^{-1}\in H if and only if there exist hHh\in H such that ba1=hb*a^{-1}=h, i.e. b=haHab=h*a\in Ha, we conclude that bRabRa iff bb is an element of Ha={hahH}H*a=\{h*a|h\in H\}. Therefore, [a]R={bGbRa}=Ha={hahH}[a]_R=\{b\in G|bRa\}=H*a=\{h*a|h\in H\}.


Consider subgroup H=(13),(14)={(1),(13),(14),(34),(134),(143)}H=\langle{(13),(14)}\rangle=\{(1),(13),(14),(34),(134),(143)\} of the symmetric group S4S_4. Since S4=24|S_4|=24 , there exist 246=4\frac{24}{6}=4 different equivalence classes S4S_4 under RR. Let us find them:


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


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