automorphism proof..

[trickslapper]

For each a in the group G, define a mapping t[sub:8pita9ce]a[/sub:8pita9ce]: G--->G by t[sub:8pita9ce]a[/sub:8pita9ce](x) = axa[sup:8pita9ce]-1[/sup:8pita9ce]. Prove that t[sub:8pita9ce]a[/sub:8pita9ce] is an automorphism of G.

I'm still reading over this chapter but i'm pretty sure i'm still gonna need help on this one. Any ideas?

 

[daon]

If f:G->G is an automorphism, it is a one-to-one and onto function from G to itself that preserves the operation in G.

1. Show that f(ab)=f(a)f(b)
2. Show that if f(a) = f(b) then a=b.
3. Show that for every y in G, there is an x in G such that f(x)=y.

 

[trickslapper]

Ok, its been a while (working on a major java program project) anyways for this proof this is what i did using your advice:

t[sub:gnsvgfd7]a[/sub:gnsvgfd7](x) t[sub:gnsvgfd7]a[/sub:gnsvgfd7](y)=(ax[sup:gnsvgfd7]-1[/sup:gnsvgfd7]a)(aya[sup:gnsvgfd7]-1[/sup:gnsvgfd7])

=axya[sup:gnsvgfd7]-1[/sup:gnsvgfd7]
=t[sub:gnsvgfd7]a[/sub:gnsvgfd7](xy)

so the operation is preserved.

Onto:

Let x exist in G, then a[sup:gnsvgfd7]-1[/sup:gnsvgfd7]xa exists in G, and t[sub:gnsvgfd7]a[/sub:gnsvgfd7](a[sup:gnsvgfd7]-1[/sup:gnsvgfd7]xa)=a(a[sup:gnsvgfd7]-1[/sup:gnsvgfd7]xa)a[sup:gnsvgfd7]-1[/sup:gnsvgfd7]

after reassociating: =(aa[sup:gnsvgfd7]-1[/sup:gnsvgfd7])x(aa[sup:gnsvgfd7]-1[/sup:gnsvgfd7])=x, therefore it is onto

One-to-One:

Let t[sub:gnsvgfd7]a[/sub:gnsvgfd7](x[sub:gnsvgfd7]1[/sub:gnsvgfd7])=t[sub:gnsvgfd7]a[/sub:gnsvgfd7](x[sub:gnsvgfd7]2[/sub:gnsvgfd7]). Now show that x[sub:gnsvgfd7]1[/sub:gnsvgfd7]=x[sub:gnsvgfd7]2[/sub:gnsvgfd7].

well t[sub:gnsvgfd7]a[/sub:gnsvgfd7](x[sub:gnsvgfd7]1[/sub:gnsvgfd7])=t[sub:gnsvgfd7]a[/sub:gnsvgfd7](x[sub:gnsvgfd7]2[/sub:gnsvgfd7]) implies that ax[sub:gnsvgfd7]1[/sub:gnsvgfd7]a[sup:gnsvgfd7]-1[/sup:gnsvgfd7]=ax[sub:gnsvgfd7]2[/sub:gnsvgfd7]a[sup:gnsvgfd7]-1[/sup:gnsvgfd7], which implies that x[sub:gnsvgfd7]1[/sub:gnsvgfd7]=x[sub:gnsvgfd7]2[/sub:gnsvgfd7], after some multiplying by a[sup:gnsvgfd7]-1[/sup:gnsvgfd7] in the back and a in the front.

therefore it is one-to-one

and since all three requirements are met t[sub:gnsvgfd7]a[/sub:gnsvgfd7] is an automorphism