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$\begingroup$

I have a subset $SB_{4}$ of $S_4$ given by

$SB_{4} = \{ id_{[1,4]},(1,2,3),(1,2,4),(1,3,2),(1,3,4),(1,4,2),(1,4,3),(2,3,4),(2,4,3),(1,2)(3,4),(1,3)(2,4),(1,4)(2,3)\} $

Question: Is $SB_{4}$ a valid subgroup of $S_4$?

neutral Element: $id_{[1,4]}$

inverse Element: $id_{[1,4]}^{-1}=id_{[1,4]}, (1,2,3)^{-1}=(3,2,1)$

I think this is not a subgroup because $(3,2,1) \notin SB_{4}$.

But the task say also: "hint: signum"

How can I use signum to define if $SB_{4}$ is really a subgroup of $S_4$?

Is there a way to check for closure without trying each possible composition of elements of $SB_{4}$?

  • 1
    $(3,2,1)$ is same as $(1,3,2)$ – 2017-02-09
  • 2
    You are wrong, $(3,2,1)$ is in $SB_4$. Check again! It is a subgroup, and is usually called $A_4$, the alternating group of degree $4$. – 2017-02-09

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Whenever you have a group homomorphism $f : G \to G_1,$ the kernel $$ \mathrm{ker}(f)=\{ x \in G : f(x)=e\} $$ is a subgroup of $G.$ Indeed, if $x_1,x_2 \in \mathrm{ker}(f),$ then $$ f(x_1 x_2)=f(x_1)f(x_2)=e e=e \Rightarrow x_1 x_2 \in \mathrm{ker}(f), $$ and if $x \in \mathrm{ker}(f),$ then $$ f(x^{-1})=(f(x))^{-1}=e^{-1}=e \Rightarrow x^{-1} \in \mathrm{ker}(f). $$ You have a particular situation of this type with $$ G=S_n \text{ and } A_n=\{ \sigma \in S_n : \mathrm{sign}(\sigma )=1\} $$ (your $\mathrm{SB}_4$ is the set $A_4$ of all even permutations of $S_4$), and, and so $A_n$ is the kernel of the homomorphism $$ \sigma \mapsto \mathrm{sign}(\sigma) $$ from $G=S_n$ into $G_1=\{-1,1\},$ and hence a subgroup of $S_n.$

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    $e$ is the neutral element? – 2017-02-09
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    It certainly is. – 2017-02-09