Question

Question:Let G be a group and Let AutG be as in Exercise 36. Let InnG be the set of all inner automorphism of G (that is, isomorphism of the form f (a) =c^-1 for some $c\in G$ , as in example 9.). Prove that InnG is a subgroup of AutG. two different elements of G may induce the same inner automorphism, that is we have c^-1ac =d^-1ad for all $a\in G$. Hence$\text{|innG|≤|G|}$,

Short answer

Answer

InnG is a subgroup of AutG.

Step-by-step solution

Required Theorem 

$$\begin{gathered} \mathbf{\text{Anon-emptysubsetHofagroupGisasubgroupofGprovidedthat,}} \\ \mathbf{\text{(i)ifa,b∈H,thenab∈H;and}} \\ \mathbf{\left(}\mathit{i}\mathit{i}\mathbf{\right)}{\mathbf{\text{ifa∈H,thena}}}^{\mathbf{-}\mathbf{1}}\mathbf{\text{∈H}} \end{gathered}$$

Proving that InnG is a subgroup of AutG

Let f (a) =c^-1 xc and g (x) =d^-1 xd be two arbitrary elements in InnG

From the composition of the group, we know that,

fog (x) = f (g(x))

= c^-1 (d^-1 xd )c

= (cd)^-1 x(cd)

Therefore,$cd\in innG$

As we assume that is an arbitrary element in InnG, and since f is a bijection from the property of bijection, we know,

role="math" localid="1651728426392" $$\begin{gathered} f^{-1}\left(x\right)={\left(c^{-1}xc\right)}^{-1} \\ =cx{c}^{-1} \end{gathered}$$

Therefore, $f^{-1}\in G$

Hence, it is proved that InnG is a subgroup of AutG