$S_6$ being the set of permutations on $6$ elements, I have to prove that there exists only one order $3$ subgroup of the group that it forms with composition.
Of course, the identity has to be part of the subgroup. In this case, let's say the subgroup is {$\boldsymbol{\alpha}_1$, $\boldsymbol{\alpha}_2$, e}. The inverses of $\boldsymbol{\alpha}_1$ and $\boldsymbol{\alpha}_2$ have to be in the subgroup, which means that either $\boldsymbol{\alpha}_1\circ\boldsymbol{\alpha}_2=\boldsymbol{\alpha}_2\circ\boldsymbol{\alpha}_1=e$, or $\boldsymbol{\alpha}_1\circ\boldsymbol{\alpha}_1=\boldsymbol{\alpha}_2\circ\boldsymbol{\alpha}_2=e$
For the second option, $\boldsymbol{\alpha}_1\circ\boldsymbol{\alpha}_2 ,\boldsymbol{\alpha}_2\circ\boldsymbol{\alpha}_1$ will also have to be a part of the subgroup. Them being equal to $e$ woudn't work since we already set inverses for them both.
I don't know how to proceed from here. Any help will be greatly appreciated!
If a group has order $3$ then, by Lagrange's theorem, the order of each of its elements is either $1$ or $3$. Of course, the only element whose order is $1$ is the identity. So, a group of order $3$ has one element whose order is $1$ and two elements whose order is $3$. But $S_3$ has exactly two elements whose order is $3$: $(1\ \ 2\ \ 3)$ and $(1\ \ 3\ \ 2)$. Since$$\bigl\{\operatorname{id},(1\ \ 2\ \ 3),(1\ \ 3\ \ 2)\bigr\}$$is a subgroup of $S_3$ with three elements, it must be the only such subgroup of $S_3$.