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I am bit confused about a fact. Let $K_4$ be the Klein $4$-group. Then it has four elements. The following fact is well-known:

The set of all permutations of $n$ objects forms a group $S_n$ of order $n!$.

So, can we say that the set of all permutations of the elements of $K_4$, forms a group $S_4$?

user_1729
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  • You can but you have to treat them as elements of a set without any group structure since the operation is completely different. – CyclotomicField Nov 11 '23 at 11:44
  • but it is said that we can create permutation group of order $n!$ from any set with $n$ elements. For a moment, if we treat $K_4$ as only a set with four elements, then it's group of permutations must be of order $4!$. By Cayley theorem every group is isomorphic to a subgroup of $S_n$. So, why we can't consider it as a group? – user_1729 Nov 11 '23 at 11:50
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    Do you perhaps mean the automorphism group of $K_4$? – Dietrich Burde Nov 11 '23 at 11:50
  • No No not automorphism group. That's $S_3$ I know. Everywhere, I see it is only mentioned about automorphisms. I think the reasoning given by @colt_browning below, may be a good one. – user_1729 Nov 11 '23 at 11:51
  • There was a similar question here, so I thought it might be relevant. – Dietrich Burde Nov 11 '23 at 11:53
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    Thanks for the link. But i already have seen that. I think in that case, the questioner had a misconception between permutation group and automorphism group. – user_1729 Nov 11 '23 at 11:56
  • You can turn any set into a symmetric group by applying permutations to the set. That doesn't have anything to do with Cayley's theorem however since the group operation in the Klein group will generally be incompatible with the group operation of the symmetric group. There does exist an injection from $K_4$ to $S_4$ but it won't be the entire group $S_4$ – CyclotomicField Nov 11 '23 at 12:59
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    @CyclotomicField actually i overlooked your comment at first glance, now i understand what you wanted to say first. That is, while thinking about all permutations of $K_4$, i have to forget about it's inner structure. So, it basically becomes an ordinary set with four elements for which the result trivially holds. – user_1729 Nov 11 '23 at 13:17

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Technically, yes. But if you speak about a group, then you are probably interested only in the permutations which are group homomorphisms. Such permutations form a strictly smaller group.

colt_browning
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