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6. ISOMORPHISMS
Problem (Page 145 # 45). Suppose g and h induce the same inner automorphism of a group G. Prove h 1g 2 Z(G).
Proof. Let x 2 G. Then
g (x)
=
h (x)
=) gxg
1
= hxh
1
=) h 1gxg 1h = x =)
(h 1g)x(h 1g) 1h = x =) (h 1g)x = x(h 1g) =)
h 1g 2 Z(G) since x is arbitrary.
Definition.
Aut(G) = { |
⇤
is an automorphism of G} and
Inn(G) = { | is an inner automorphism of G}.
Theorem (6.4 — Aut(G) and Inn(G) are groups).
Aut(G) and Inn(G) are groups under function composition.
Proof.
From the transitive portion of Theorem 3, compositions of isomorphisms are
isomorphisms. Thus Aut(G) is closed under composition. We know function
composition is associative and that the identity map is the identity automorphism.
Suppose ↵ 2 Aut(G). ↵
1
is clearly 1–1 and onto. Now
↵ 1(xy) = ↵ 1(x)↵ 1(y) () ((= by 1–1)
⇥
⇤
⇥
⇤
⇥
⇤ ⇥
⇤
↵ ↵ 1(xy) = ↵ ↵ 1(x)↵ 1(y) () xy = ↵ ↵ 1(x) ↵ ↵ 1(y) ()
xy = xy.
Thus ↵
1
is operation preserving, and Aut(G) is a group.