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Let K be the set of all permutations of $ S_4 $ type $ [2 ^ 2] $ and the identity permutation $\in K $. Prove that $ K $ is a subgroup $ S_4 $. I would like to prove that for$\pi, p \in S_4$ and type of $[2^2]$
$\pi p $ also is in K. But I can't deal with it.

user180834
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Of the $24$ permutations in $S_4$ there are exactly $3$ permutations of the desired type. Write them explicitly, and then verify that if you multiply two of them you get the third.

The above shows that $K$ is closed under multiplication, to finish the argument that $K$ is a subgroup show that each of your $3$ non-trivial elements is its own inverse and thus $K$ is also closed under taking inverses.

DKal
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  • yes, it's one solution but solve it in general way because the number can be much bigger. – user180834 Oct 23 '14 at 17:10
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    What do you mean 'in a general way'? Your questions asks about $S_4$. – DKal Oct 23 '14 at 17:13
  • ok, but what if my question will be about $S_{99}$ – user180834 Oct 23 '14 at 17:15
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    Then that'd be another question @user180834...and then the set of all permutations of type $;[2^2];$ is not even closed, and thus not a subgroup: $$(12)(34)\cdot(12)(35)=(354)$$ – Timbuc Oct 23 '14 at 17:23
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    In that case, the claim is not true. For example take $(1,2)(3,4)$ and $(5,6)(7,8)$. The above two permutations are of type $2^2$, but their product is of type $2^4$. – DKal Oct 23 '14 at 17:24
  • help... I wrote 3 permutations: $(1 2)(1 3) , (1 2 ) (1 4), (1 3)(1 4)$ And then I compose $(1 2 )(1 3 )(1 2 ) (1 4 ) and multiply them doesn't give third. :( – user180834 Oct 23 '14 at 18:17
  • The permutation $(1,2)(1,3)$ is not of type $2^2$, it is of type $3^1$ because in disjoint cycle notation it is $(1,2,3)$. – DKal Oct 23 '14 at 18:48