Abstract Algebra - automorphism

nebbish
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I have two problems I would like to discuss.

1.For any group G prove that the set of inner automorphisms J(G) is a normal subgroup of the set of automorphisms A(G).

Let A be an automorphism of G. Let T_{g} be an inner automorphism, i.e.
xT_{g}=g^{-1}xg

The problem can be reduced to the question whether the following equality is true:
xAT_{g}=xT_{g}A

Then expanding using xT_{g}=g^{-1}xg we have:
gxAg^{-1}=gxg^{-1}A

However I am having trouble proving this equality. Attempting to use the definition of normal more directly also did not work, i.e. showing that
AT_{g}A^{-1} is in J(G).

2.Let G = {e, a, b, ab} where a^2=b^2=e and ab = ba. Determine the set of automorphisms A(G).
This problem could be handled easily using brute force but I would like some way to narrow down the possible automorphisms. My attempts to solve the problem so far have resulted in way too many calculations.
 
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2. Really, way to many calculations? There are only 6 possible permutations of a,b,ab to consider.1. If you are still strugglng with this, what inner automorphism do you think you ought to get for AT_gA^{-1}? It is T_h for some h, try taking a reasonable guess as to what h ought to be.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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