Generating the symmetric group $S_n$

I know that $\sigma =(1 2 \ldots n)$ and $\tau =(1 2)$ together should generate the symmetric group by virtue of conjugation, i.e.

$(\sigma)^k \circ \tau \circ (\sigma^{-1})^k = (k+1, k+2)$; we know that the set of adjacent transpositions generates $S_n$, so we're done.

However-- and I realize that this question is incredibly dumb-- when I try this for $n=3$, I should have that $\sigma \circ \tau \circ \sigma^{-1} = (2 3)$. But I get

$\sigma \circ \tau = (1 2 3) \circ (1 2) = (2 3)$ when I work it out by hand, writing out the digits and physically switching them; then $(2 3)\circ \sigma^{-1}= (1 3)$. What am I doing wrong?

EDIT: I'd confused sigma and tau. I've edited the question to reflect this.

Solution 1:

Please NOTE: the answer below was written based on the original post: $$\sigma = (12), \;\tau = (123 \cdots n)$$

The original post has now been edited.

When composing $\sigma\circ \tau \circ \sigma^{-1}$, we can first compose $\alpha = \tau \circ \sigma^{-1}$, and then compose $\sigma(\alpha)$. (Or else first compose $\sigma \circ \tau = \beta,$ and then compose $\beta(\sigma^{-1}).$

In your case $\tau\circ \sigma^{-1} = (123)(12) = (13)$, and then $\sigma\circ(\tau\circ \sigma^{-1}) = (12)(13) = (132).$

And as Daniel Fischer points out, you seem to have confused yourself when you conclude