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Let $K=\{ (1),(12)(34), (13)(24),(14)(23)\}$

(a)show that it is a subgroup of $A_4$ and hence of $S_4$

(b) state number of cosets in K $in $ $A_4$

c) state number of cosets of $K$ in $S_4$ dont list them

comfortable with permutation matrix. Not sure what $A_4$ means but the permutation the elements. feel that can show it is a subgroup by having to do a 4 by 4 table.

Attempt 1

still multiplying out but its a bad Idea $$\begin{aligned} \\(1)*(1)&=(1) \\(1) * (12)(34)&=(12)(34) \\(1)* (23)(24)&=(13)(24) \\(1)* (4)(23)&=(14)(23) \end{aligned} $$

$$\begin{aligned} \\(12)(34) * (12)(34)&= \\(12)(34)* (23)(24)&= \\(12)(34)* (4)(23)&= \end{aligned} $$

1 Answers 1

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a) You're correct in trying to multiply everything out, though there might just be a shorter way of doing it. To show it's a subgroup of $A_4$ you also need to say why each element is in $A_4$ though.

b) Lagrange's theorem is the way to go here.

c) Again, Lagrange's theorem will help you out.

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    I know that $|S_4|=!4=4*3*2*1=24$ so for eeach coset it depends of each element in the group how many itself geenrates and dived it by 24 to find the lements of each coset . Not sure what will be the case of $A_4$ – 2017-02-22
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    Careful with the notation, while you're correct that $|S_4| = 24$, the notation $!4$ refers to something different than $4 \times 3 \times 2 \times 1$, the correct notation is $4!$. – 2017-02-22