Show that if H is any group then there is a group G that contains H as a normal subgroup with the property that for every automorphism f of H there is an element g of G such that the conjugation by g when restricted to H is the given automorphism f, i

Proof:
We have an identity map id from Aut(H) to itself, such that id(f)=f
for any f in Aut(H). We define G=H x_id Aut(H).
Then for any (h1,f1), (h2,f2) in G, we have
(h1,f1)*(h2,f2)=(h1*id(f1)(h2),f1f2)=(h1f1(h2),f1f2).
H and Aut(H) can be identified as subgroup of G.
1. I want to show that H is a normal subgroup of G. Let e be identity of Aut(H).
For any (h,f) in G and (h',e) in H, we can verify that
(h,f)^(-1)=(f^(-1)(h^(-1)),f^(-1))
then we have
(h,f)*(h',e)*(h,f)^(-1)
=(hf(h'),f)*(f^(-1)(h^(-1)),f^(-1))
=(hf(h')*f(f^(-1)(h^(-1))),ff^(-1))
=(hf(h')h^(-1),e)
which belongs to H.
Thus H is normal in G
2. For every f in Aut(H), we set g=(e_H,f) in G. We restrict g on H, then for
every (h',e) in H, we make a conjugate of g on (h',e) and get
(e_H,f)*(h',e)*(e_H,f)^(-1)
=(e_H f(h') e_H^(-1), e)
=(f(h'),e)
=f(h',e)
So every isomorphism f in Aut(H) can be obtained as an inner isomorphism of G
restricted to H.

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