Automorphisms

**mpryal** · April 22nd 2009, 10:03 PM

Let a,b be automorphisms of G, and let ab be the product of a and b as mappings on G. Prove that ab is an automorphism of G, and that the inverse of a is an automorphism of G, so that the set of all automorpisms of G is itself a group.

**Gamma** · April 22nd 2009, 10:41 PM

Are you sure you want your group operation in Aut(G) to be pointwise multiplication? I think you would want composition to be the group operation.
For instance, let $\sigma^1, \sigma^2 \in Aut(G)$ . And consider the following.

$\sigma^1(xy)\sigma^2(xy)=\sigma^1(x)\sigma^1(y)\si gma^2(x)\sigma^2(y)$

And what you want to say is that this should be
$\sigma^1(x)\sigma^2(x)\sigma^1(y)\sigma^2(y)$

But this requires G to be commutative since you need to switch the two inner terms if it is to be a homomorphism.

If instead you use composition
$\sigma^1(\sigma^2(xy))=\sigma^1(\sigma^2(x)\sigma^ 2(y))=\sigma^1(\sigma^2(x))\sigma^1(\sigma^2(y))$ which is a homomorphism. Furthermore the composition of bijections is a bijection, so it is an isomorphism and clearly it still goes from G to G so it is an automorphism.

Inverses are particularly clear for function composition as $\sigma \in Aut(G)$ necessarily means it is an isomorphism from G to G and therefore has an inverse that is an isomorphism $\exists \sigma^{-1}\in Aut(G)$ , this clearly goes from G to G in the oposite direction and gives you the identity map on G. That is $\sigma(\sigma^{-1}(x)=Id(x)=\sigma^{-1}(\sigma(x))$

Associativity and Identity are clear. (composition of functions is associative and the identity is clearly an automorphism from G to G.)

I hope this helps.

**aliceinwonderland** · April 22nd 2009, 10:52 PM

Originally Posted by mpryal

Let a,b be automorphisms of G, and let ab be the product of a and b as mappings on G. Prove that ab is an automorphism of G, and that the inverse of a is an automorphism of G, so that the set of all automorpisms of G is itself a group.

Let a be an automorphism of G such that $a:G \rightarrow G$ defined by $a(g) = \sigma_a \cdot g$ where $\sigma_a,g \in G$ ; let b be an automorphism of G such that $b:G \rightarrow G$ defined by $b(g) = \sigma_b \cdot g$ where $\sigma_b,g \in G$ .

Let $ab:G \rightarrow G$ defined by $ab(g) = \sigma_{ab} \cdot g$ where $\sigma_{ab},g \in G$ and $\sigma_{ab}=\sigma_{a}\cdot \sigma_{b}$ . If $g_1 = g_2$ , then $ab(g_1)=\sigma_{ab}(g_1) = \sigma_{a}(\sigma_{b}(g_1))=\sigma_{a}(\sigma_{b}( g_2))=ab(g_2)$ for $g_1, g_2 \in G$ . Thus, $ab$ is well-defined and it is an automorphism since $(ab)(ab)^{-1} = 1_{G}$ and $(ab)^{-1}(ab) = 1_{G}$ .

Let $a^{-1}:G \rightarrow G$ defined by $a^{-1}(g) = \sigma_a^{-1} \cdot g$ where $\sigma_a^{-1}, g \in G$ . Since a is an automorphism, $a^{-1}$ is clearly automorphism.

Now it remains to show that the set of all automorphisms of G itself a group. The identity element of this group is $1_{G}$ , which maps an element of G to itself. As shown above, the inverse of an automorphism of a is $a^{-1}$ . It is left to you to show the associative law holds for this group.

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