Definition of Automorphism

Question

I really did not get the idea of automorphism. I actually cannot find a good source(textbook or lecture video) in order to understand the automorphism. I have seen lots of definition, I think I need to be shown practices. Can someone show me an example about automorphism? later I will try to do the same on my own.

Answer 1

First, we define group homomorphism to be a map from one group to another: $f:G\rightarrow H$ that preserve the group structure, i.e. $f(e)=e'$ and $f(g_1)f(g_2)=f(g_1g_2)$ where $e$ is identity in $G$ and $e'$ is identity in $H$.

Then group isomorphism is defined to be bijective homorphism, and it can be proved that the inverse of isomorphism is also homomorphism.

Further, if the domain and codomain of an isomorphism is the same, then the isomorphism is called an automophism. In ohter word, automorphism is isomorphism from one group to itself.

An important kind of isomorphism is conjugation: for any $g\in G$, a conjugation by $g$ is the following automorphism: $c_g:G\rightarrow G,x\mapsto gxg^{-1}$.

Answer 2

One way to understand a definition is by examples. Just to get "the idea of an automorphism ". The smallest non-abelian group is the symmetric group $S_3$ of order $6$, just to take some popular example. It has the elements $$ (1),(12),(13),(23),(123),(132). $$ The transpositions have order $2$ and the $3$-cycles have order $3$. Any automorphism has to preserve these orders (note that $f(a^n)=f(a)^n$ for any automorphism). This makes it easy to find all automorphisms of $S_3$:

Finding the automorphisms of $S_3$ by looking at the orders of the elements

Answer 3

Definition by category theory

Let $C$ be a category and $A, B \in obj(C)$. Then morphism $f: A\to B$ is Isomorphism when there exists morphism $g: B\to A$ such that $fg=id_{B}$ and $gf=id_{A}$.

Now Automorphism $h$ on $X\in obj(C) $ is an isomorphism on $X$.

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For example in category $\mathbb{G}rp$, if $G$ is arbitrary group, then homomorphism $f: G\to G$ is Automorphism iff there exists $g=G\to G$ such that $fg=gf=id_{G}$.