For a fixed group G, prove that the set of all automorphisms of G forms a group

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Explanation / Answer

Let Aut(G) be the set of all automorphisms ?: G --> G. In order to show that this is a group under the operation of composition, we must verify: (1) Is the set is closed under composition? Yes! If you are given isomorphisms ?, ?: G --> G, then it is not too tough to show that ??? and ??? are isomorphisms. I can expand on this in more detail if you like, but you have probably seen a proof before that a composition of bijective functions is bijective. If a and b are elements of the group, ???(ab) = ?(?(ab)) = ?(?(a)?(b)), because ? is an isomorphism. Since ? is also an isomorphism, ?(?(a)?(b)) = ???(a)???(b), so the composition ??? preserves products. Thus, ??? is an isomorphism if ? and ? are. (2) Is the set associative? Yes! All you need to do is show that, for any three isomorphisms ?, ? and ?, ??(???) = (???)??. To do that, just show that for each x in G, ??(???)(x) = (???)??(x) = ?(?(?(x))). It's just pushing around definitions. (3) Does the set contain an identity element? Yes! Let the identity automorphism e: G --> G be the map e(x) = x. Clearly, e?? = ??e = ?. (4) Does each element of the set have an inverse under ?? Yes! Since each isomorphism ?: G --> G is bijective, there is a well-defined inverse map ?^(-1): G --> G. You may have already seen a proof that the inverse of an isomorphism is an isomorphism. If not, it isn't too difficult to prove: I'll leave it to you, but I can expand on it if you need me to. Further, the composition ?^(-1) ? ? = ? ? ?^(-1) = e. Since Aut(G) satisfies all the group axioms, it forms a group under ?, as needed.

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