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My Question Reads:

Let G be a group and let AutG be the set of all automorphisms of G. Also, let InnG be the set of all inner automorphisms of G. Prove that InnG is a subgroup of AutG.

My thoughts:

I believe I would need to first state the definitions of an automorphism which is a function from a group to itself and an inner automorphism is where again a function from G to itself occurs but where f(a)=c^-1*a*c.

From here I am not too sure if I need to then prove closure and inverse to confirm subgroup but I am confused how to do this considering these are the type of functions I am dealing with.

For closure, would I need to pick two inner automorphisms or am I still picking two values of G?

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    you can start with existence of identity, then closure and inverse should seal the deal for the subgroup part. For the Inn(G) you can do the same thing, rather show that for any two automorphisms in Inn(G) $\phi$ and $\psi$, $\phi$$\psi^{-1}$ is also in Inn(G) – 2017-02-11
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    @HumbleStudent so it would not be enough to show it is a subgroup? I thought the question asked to show just that InnG is a subgroup of AutG – 2017-02-12
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    Oh.. I apologize. I thought the question asked both to show Aut(G) is a group and Inn(G) is the subgroup. Well if it only the second, you don't really need the first part. – 2017-02-12
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    @HumbleStudent Oh sorry if my wording is weird but this is how the problem is stated. Then I only have to show InnG is a subgroup, but I am lost as to how to start. I see you said to choose two functions in InnG but then am I just showing inverse? – 2017-02-13
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    closure and inverse. So Let us say that you have $\phi_{x}$ and $\phi_{y}$ , two elements of Inn(G). Then you have to show that multiplication of these two elements is still in Inn(G) . Basically you have to show what element causes the inner automorphism, that results as multiplication of these two inner automorphisms – 2017-02-13

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