Let $G$ be a group of order $4$, $G$ $=$ {$e$,$a$,$b$,$ab$}, $a^2 = b^2 = e , ab = ba$. Determine $A(G)$.Where $A(G)$ is set of all automorphisms of $G$.
What i did was - Since we know that for Homomorphism $e$ should map to $e$ in this case .Hence we are left with $3!$ $=$ $6$ permutations to be evaluated and then for each of them i checked homomorphic condition i.e $Ф(xy) = Ф(x)Ф(y)$ .
I found that all of the six permutations are isomorphism and hence they belong to $A(G)$.
Is there any better method for finding $A(G)$ without evaluating and testing all the cases ( i mean just by little observation can we solve this problem ) ?
You have Calculated there are 6 element in Aut(G). Now there are only 2 possibilities Eithere $Aut(G)\cong S_3$ or $Aut(G)\cong Z_6$
Now ,
In case It is isomorphism to $Z_6$ then there are only one element of order 2 .but
Here
1) $a\to b,b\to a$ all other elements are maps to same
2) $a\to ab,ab\to a$ all other elements are maps to same
Similarly you can contruct one more element which have order2. SO
$Aut(G)\cong S_3$