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 December 10th, 2010, 06:27 PM #1 Member   Joined: Oct 2009 Posts: 85 Thanks: 0 Permutation proof Let $A$ be a set and $a \in A$. Let $G$ be the subset of $S_A$ consisting of all the permutations $f$ of $A$ such that $f(a)= a$. Prove that $G$ is a subgroup of $S_A$. Clueless...
 December 10th, 2010, 08:07 PM #2 Senior Member   Joined: Nov 2010 Posts: 502 Thanks: 0 Re: Permutation proof What happens when you multiply two of them together? Or if you take three of them? Does there exist an inverse? (yes) Why?
 December 12th, 2010, 05:18 PM #3 Member   Joined: Oct 2009 Posts: 85 Thanks: 0 Re: Permutation proof So in G, everthing maps everything to itself. H is a subset of G so that covers that part of a subgroup. Should I prove closure and inverses using composition (as opposed to multiplication) since my book does not even mention the multiplication of permutations but only talks about composition of permutations and the group of all the permutations of A is S_A. So the composition of two functions that mapped to themselves will be in G, but I have no idea how to prove this. Here's what I am unsure about: Is it sufficient enough just to say that since the composite of any two permutations of A is a permutation of A that $f(a) \circ f(b) \in G$ and is it also sufficient enough to say that by definition the inverse of any permutation of A is a permutation of A therefore $f^{-1}(a) \in G$. Then we have all the conditions for a subgroup. Is this ok to do?

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