How to obtain the automorphism group of $D_2$? I'm a new learner on Group Theory, using a book for physics students.
When trying to show the definition of semi-direct product, the author of the book give the following equation:
$$Aut\ D_2=S_3.$$
I know what $D_2$ and $S_3$ are, and I kind of know that $Aut\ D_2$ means that the set of all of $D_2$'s automorphisms combined with composition as multiplication forms a group called $Aut\ D_2$.
But I don't know how to get the answer.
Thank you in advance!
Cheers,
Collin
(P.S.:  I'm a physics student,... So.. ah..,   please involve as little new concepts as possible, I would really appreciate that.)
 A: An automorphism must fix the identity element. 
To show that any permutation of the remaining three elements in the Klein is an automorphism, it is enough to check that any transposition (of two non-identity elements) preserves multiplication. 
That is an easy calculation. 
A: Note that $D_2$ is the product $C_2\times C_2$, and you can check that a group automorphism of this is the same as a linear automorphism over $C_2$, which happens to be a field, too. Now this is the group $\mathsf{GL}(2,2)$, which one knows (from doing linear algebra over finite fields, of course) has size $(2^2-2)(2^2-1)=6$. Since it is not abelian, it must be $S_3$, the only non-abelian group of order $6$.
A: $
\newcommand{\Aut}{\operatorname{Aut}}%
$Here's a more explicit correspondence.  Let $D_2$ have identity element $e$, and be generated by $a$ and $b$ with $a^2 = b^2 = e$ and $ab=ba$.  Any two of $a$, $b$ and $ab$ will generate the whole group.
An automorphism $\lambda$ of $D_2$ is determined by where it sends the generators.  There are three choices for $\lambda(a)$ and two remaining for $\lambda(b)$.  So $|\Aut D_2| = 6$.
We can enumerate them:
$$
    \begin{array}{c|cc} \lambda & \lambda(a) & \lambda(b) & \lambda(ab) 
    \\\hline
        \iota & a & b & ab 
    \\
        \rho  & b & ab & a 
    \\
        \rho^2  & ab & a & b 
    \\
         \phi   & b  & a & ab 
    \\
         \rho\phi   & ab  & b & a 
    \\
         \rho^2\phi   & a  & ab & b 
    \end{array}
$$
Notice that $\rho^3 = \phi^2 = \iota$, and $\phi \rho \phi = \rho^{-1}$.  This characterizes $S_3$.  If you think of $a$, $b$, and $ab$ as the vertices of an equilateral triangle, $\phi$ is a one-third rotation, $\phi^2$ is a two-thirds rotation, or one-third in the other direction, and each of $\phi$, $\rho\phi$, and $\rho^2\phi$ swap vertices.
