Show that the subgroup $\langle\tau\rangle=\{\tau^n | n \in \mathbb{Z}\}$ is normal in $S_3$ where $\tau \in S_3$ is the 3-cycle:
$(123) = \begin{bmatrix}1&2&3\\2&3&1\end{bmatrix}$.
So for $g\in S_3$, $h\in \langle\tau\rangle$, I have to prove that $ghg^{-1}\in \langle\tau\rangle$ or equivalently $g\langle\tau\rangle g^{-1} = \langle\tau\rangle$. I have no idea how to do this. I think the inverse would be $(312)$, so I have to show $(123)*\tau^n*(312)=\tau^n$?
No, you can check that $(123)(312) \neq 1$. In fact, $(312) = (123)$ and the square of a three-cycle cannot be the identity.
Instead, you can check that $(123)^{-1} = (132)$. (Which is also equal to $(321)$ and $(213)$).
But I'm not sure why you were trying to compute its inverse anyway.
No, you need to show that given any $n \in \Bbb Z$ and $\sigma \in S_3$, there exists $m \in \Bbb Z$ such that $$\sigma \tau^{\color{red}{n}} \sigma^{-1} = \tau^{\color{red}{m}}.$$
Moreover, since $\tau$ generates the subgroup $\langle \tau \rangle$, it suffices this to prove for this $n = 1$ (and all $\sigma$). (Do you see why?)
Thus, you only need to show that for every $g \in S_3$, $$g \tau g^{-1}$$ is in $\langle \tau \rangle$, i.e., is a power of $\tau$.
First note that this is clearly true if $g$ is in $\langle \tau \rangle$ as $g$ and $\tau$ commute. This takes care of three elements.
Now, suppose $\sigma \notin \langle \tau \rangle$. In this case, check that the other elements of $S_3$ are $\sigma\langle \tau \rangle = \{\sigma, \sigma \tau, \sigma \tau^2\}$.
By an argument as earlier, it suffices to prove this for $g = \sigma$ (why?).
That you can do by simply picking any $\sigma \notin \langle \tau \rangle$.