A Group Presentation for S3

# A Group Presentation for S3

Consider $S_3$, the symmetric group on $3$ elements. In cycle notation we may denote all of the elements in $S_3$ by:

(1)
\begin{align} \quad S_3 = \{ (1), (1, 2), (1, 3), (2, 3), (1, 2, 3), (1, 3, 2) \} \end{align}

Consider the following group presentation:

(2)
\begin{align} \quad \langle a, b : a^3 = 1, b^2 = 1, aba^{-2}b = 1 \rangle \end{align}

We will show that this $S_3$ has the above group presentation. Define a function $f : \langle a, b : a^3 = 1, b^2 = 1, aba^{-2}b = 1 \rangle \to G$ on the generators of $\langle a, b : a^3 = 1, b^2 = 1, aba^{-2}b = 1 \rangle$ by:

(3)
\begin{align} f(a) &= (1, 2, 3) \\ f(b) &= (1, 2) \end{align}

Then observe that:

(4)
\begin{align} \quad (1, 2, 3)^3 = (1) \quad (1, 2)^2 = (1) \end{align}

We also have that:

(5)
\begin{align} \quad (1, 2, 3)(1, 2) = (1, 2)^{-1}(1, 2, 3)^2 \end{align}

Therefore $f(a^3) = 1$, $f(b^2) = 1$, and $f(aba^{-1}b) = 1$. So the function $f$ is a well-defined surjective homomorphism. Now observe that $\langle a, b : a^3 = 1, b^2 = 1, aba^{-2}b = 1 \rangle$ has only six elements. Therefore $f$ is also injective. Hence:

(6)
\begin{align} \langle a, b : a^3 = 1, b^2 = 1, aba^{-2}b = 1 \rangle \cong S_3 \end{align}