Given the subgroup of $S_4$: $$ V_4:=\{\text{id},(12)(34),(13)(24),(14)(23) \} $$ How do I check if this is a normal subgroup? I know that to be the case it has to hold that $$ \forall\sigma\in S_4: \sigma V_4=V_4 \sigma $$ this is equivalent to $$ \sigma V_4\sigma^{-1} \subset V_4 $$ I just don't know how to check this. It would be clear if a $\sigma$ would be given, but can I just use any $\sigma$? What I also know is that $(12)(34)=(34)(12)$ that follows from trying out or using the characterization of disjunct cycles. So basicly $(12)(34)=((12)(34))^{-1}$. So for $\sigma (12)(34)=(12)(34)\sigma \Leftrightarrow (12)(34) \sigma (12)(34)=\sigma$ if we "add" $(12)(34)$ from the left right? That would work for the other elements as well. Is that correct?
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Arnaud D.
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KingDingeling
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3Note that conjugate permutations have the same cycle type – J. W. Tanner Feb 11 '20 at 16:30
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Related : https://math.stackexchange.com/questions/157876/normal-subgroups-of-s-4 – Arnaud D. Feb 11 '20 at 16:36
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1Hint: Conjugation in $S_n$ respects cycle structure. Your subgroup contains all elements that are products of two disjoint transpositions. – Arturo Magidin Feb 11 '20 at 16:52