Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 4.2 Exercise 4.2.1
Solution: The multiplication table for $G$ is as follows.
1 a b c
1 1 a b c
a a 1 c b
b b c 1 a
c c b a 1
We can see that with $1,a,b,c$ labelled as 1,2,4,3, respectively, since $a1 = a$, $aa = 1$, $ab = c$, and $ac = b$ we have $a(1) = 2$, $a(2) = 1$, $a(3) = 4$, and $a(4) = 3$. Thus $a \mapsto (1\ 2)(3\ 4)$.
Similarly, $b1 = b$, $ba = c$, $bb = 1$, and $bc = a$, so that $b \mapsto (1\ 4)(2\ 3)$. Likewise $c \mapsto (1\ 3)(2\ 4)$.
Now we relabel the elements $1,a,b,c$ as 1,4,2,3, respectively. Now $a(1) = a1 = a = 4$, $a(2) = ab = c = 3$, $a(3) = ac = b = 2$, and $a(4) = aa = 1$. Thus $a \mapsto (1\ 4)(2\ 3)$. Now $b(1) = b = 2$, $b(2) = bb = 1$, $b(3) = bc = a = 4$, and $b(4) = ba = c = 3$, so that $b \mapsto (1\ 2)(3\ 4)$, and $c(1) = c = 3$, $c(2) = cb = a = 4$, $c(3) = cc = 1$, and $c(4) = ca = b = 2$, so that $c \mapsto (1\ 3)(2\ 4)$. Clearly this is the same subgroup as the image of $G$ under the first labelling.
