Show that ϕa ∈ Inn(g) is an automorphism of G.

Any help on this, I'm kinda lost.

Printable View

- Apr 1st 2012, 04:36 PMJohngalt13Proof involving automophism
Show that ϕa ∈ Inn(g) is an automorphism of G.

Any help on this, I'm kinda lost. - Apr 1st 2012, 05:49 PMDevenoRe: Proof involving automophism
what is an automorphism of G?

1. it is a homomorphism from G to G

2. it is bijective

so we have to check conditions (1) and (2) above.

to prove (1), we need to show that $\displaystyle \phi_a(xy) = \phi_a(x)\phi_a(y)$ for all x,y in G.

but $\displaystyle \phi_a(xy) = a(xy)a^{-1} = (axa^{-1})(aya^{-1}) =\dots$ can you continue?

to prove (2), we have to show 2 things as well:

2.a. $\displaystyle \phi_a$ is injective

2.b. $\displaystyle \phi_a$ is surjective

to prove (2.a) it suffices to show that $\displaystyle \mathrm{ker}(\phi_a) = \{e\}$. so suppose $\displaystyle x \in \mathrm{ker}(\phi_a)$.

this means that $\displaystyle axa^{-1} = e$, so $\displaystyle ax = a$, so what can we say about x? (what is the unique element of G such that ax = a?).

to prove (2.b) for any y in G, we need to find some x in G with $\displaystyle \phi_a(x) = y$. how about $\displaystyle x = a^{-1}ya$ ?

*******

alternately, to show $\displaystyle \phi_a$ is bijective, we could exhibit a (two-sided) inverse. what is the logical candidate for $\displaystyle (\phi_a)^{-1}$ ?

(maybe $\displaystyle \phi_{a^{-1}}$? can you prove this ?) - Apr 1st 2012, 06:11 PMJohngalt13Re: Proof involving automophism
Thanks!

I think I got it, I'm a little lost on how to handle the step where you go from ϕa(xy)=a(xy)a^-1 though still

If ϕ is given an operation I get it, but how do you know what to do when you don't really know what the mapping is?

Thanks - Apr 1st 2012, 07:58 PMDevenoRe: Proof involving automophism
by definition the inner automorphism of G induced by an element a of G, is conjugation by a:

$\displaystyle \phi_a(x) = axa^{-1}$, for all x in G.

since xy is an element of G, $\displaystyle \phi_a(xy)$ is xy conjugated by a: $\displaystyle \phi_a(xy) = a(xy)a^{-1}$.

in other words there are two groups at play here:

one group is G, whose elements are just elements of the set G.

the other group is Aut(G) whose elements are functions (special functions, that preserve the multiplication of G, and are bijective).

in G, we multiply. in Aut(G), we compose functions.

now, for any element g in G, there is an element $\displaystyle \phi_g$ in Aut(G), called the inner automorphism induced by G.

and, in fact, the mapping $\displaystyle g \mapsto \phi_g$ is a homomorphism itself of G into Aut(G).

the image of the homomorphism is a subgroup of Aut(G), called the group of inner automorphisms of G, or Inn(G).

the thing is, this usually ISN'T an isomorphism, because two different elements of G might give the same automorphism.

let's look at an example:

suppose $\displaystyle G = D_4 = \{1,r,r^2,r^3,s,rs,r^2s,r^3s\}$ where $\displaystyle r^4 = s^2 = 1, sr = r^{-1}s$.

let's compute the different automorphisms $\displaystyle \phi_g$.

first of all, it's clear that $\displaystyle \phi_1(g) = 1g1^{-1} = 1g1 = g$ for all $\displaystyle g \in D_4$.

this is a special automorphism, the identity automorphism, which is the identity of Inn(G) (and of Aut(G), as well).

you may take my word for it (or verify it yourself) that $\displaystyle r^2$ commutes with all of $\displaystyle D_4$.

so $\displaystyle \phi_{r^2}(g) = r^2gr^{-2} = gr^2r^{-2} = g1 = g$, so $\displaystyle \phi_{r^2} = \phi_1$ even though $\displaystyle 1 \neq r^2$.

next, let's look at $\displaystyle \phi_r$.

$\displaystyle \phi_r(1) = r1r^{-1} = rr^{-1} = 1$ (duh!)

$\displaystyle \phi_r(r^k) = r(r^k)r^{-1} = r^{1+k-1} = r^k$ so $\displaystyle \phi_r$ is the identity on $\displaystyle \langle r \rangle$.

$\displaystyle \phi_r(s) = rsr^{-1} = rsr^3 = r(sr)r^2 = r(r^{-1}s)r^2 = sr^2 = (sr)r = (r^{-1}s)r $

$\displaystyle = r^{-1}(sr) = r^{-1}(r^{-1}s) = r^{-2}s = r^2s$

$\displaystyle \phi_r(r^ks) = r(r^ks)r^{-1} = (r^k)(rsr^{-1}) = r^{k+2 (mod\ 4)}s$

(so for example, $\displaystyle \phi_r(r^3s) = r^5s = rs$).

i will leave it to you to prove that $\displaystyle \phi_r = \phi_{r^3}$.

next we look at $\displaystyle \phi_s$:

$\displaystyle \phi_s(1) = 1$ (easy)

$\displaystyle \phi_s(r) = srs^{-1} = (sr)s = (r^{-1}s)s = r^3s^2 = r^31 = r^3$

$\displaystyle \phi_s(r^2) = sr^2s = (srs)(srs) = r^6 = r^2$

$\displaystyle \phi_s(r^3) = sr^3s = (srs)(srs)(srs) = r^9 = r$

$\displaystyle \phi_s(s) = s^3 = s$

$\displaystyle \phi_s(rs) = s(rs)s = srs^2 = sr = r^{-1}s = r^3s$

$\displaystyle \phi_s(r^2s) = s(r^2s)s = sr^2 = r^2s$

$\displaystyle \phi_s(r^3s) = s(r^3s)s = sr^3 = sr(r^2) = r^2(sr) = r^2(r^{-1}s) = rs$

i will state without proof that $\displaystyle \phi_s = \phi_{r^2s}$ , and that you can compute $\displaystyle \phi_{rs}$ by composing $\displaystyle \phi_r$ and $\displaystyle \phi_s$.

it follows that $\displaystyle \phi_{r^3s} = \phi_{r^2} \circ \phi_{rs} = \phi_1 \circ \phi_{rs} = \phi_{rs}$.

in fact (you can verify the details yourself, it's good practice for you): $\displaystyle \mathrm{Inn}(D_4) = \{\phi_1,\phi_r,\phi_s,\phi_{rs}\}$. - Apr 2nd 2012, 05:10 PMJohngalt13Re: Proof involving automophism
Thanks a lot! That was incredibly helpful