Test question that I got destroyed by. Thought about it all weekend and came up with an explanation. Today, scores back, and prof gave a very different explanation from the one I had in mind. But I'm wondering if my explanation could also work? I'm pretty sure the question is about a special case of inner automorphism, though we haven't covered this topic yet so neither his proof nor mine makes explicit reference to the topic. Here's the question:
Let $n \geq 3$ be an integer. Suppose that $\sigma \in S_n$ sends $a$ to $b$, where $a$ and $b$ are distinct elements of $\{1, 2, ..., n\}$. With justification, tell whether there exists a transposition $\tau \in S_n$ such that $\tau \sigma \tau \neq \sigma$.
His explanation:
Let $\tau = (b \ c)$ with $c \in \{1, 2, ..., n \} \setminus \{a, b \}$. Then $\tau \sigma \tau (a) = c$ whereas, by assumption in the question stem, $\sigma(a) = b$.
(Note that if there is an error above, it is certainly an error I made taking notes and not my prof's error!) This proof is definitely very direct. But I find it unsatisfying because it doesn't give me a lot of insight into when $\tau \sigma \tau = \sigma$ and when not. So my own explanation focuses on finding a criterion for when $\tau \sigma \tau = \sigma$, and then showing that any $\sigma \in S_n, n \geq 3$ will sometimes fail that criterion:
I claim that $\tau \sigma \tau = \sigma$ if and only if $\sigma$ and $\tau$ commute.
($\Rightarrow$) If $\tau \sigma \tau = \sigma$, then $\sigma \tau = \tau^{-1} \sigma = \tau \sigma$, where the last equality is because $\tau$ is a transposition. Therefore $\tau$ and $\sigma$ commute.
($\Leftarrow$) If $\tau$ and $\sigma$ commute, then $\tau \sigma \tau = \tau \tau \sigma = \sigma$, where again the last equality is because $\tau$ is a transposition.
However, for any $\sigma \in S_n, n \geq 3$, there exists some $\tau$ that is neither disjoint to $\sigma$ nor equal to $\sigma$ (if $\sigma$ is a transposition), and therefore there is some $\tau$ that does not commute with $\sigma$, meaning that $\tau \sigma \tau \neq \sigma$.
I think the two explanations are actually equivalent because in prof's proof, both $\tau$ and $\sigma$ move $b$, so they are not disjoint. And the criteria that $c \neq a$ means that $\tau \neq \sigma$, if $\sigma$ is a transposition. I also think mine is correct because $\tau \sigma \tau$ is an inner automorphism when $\tau$ is a transposition, and for any inner automorphism, $A B A^{-1} = B$ iff A and B commute...right?
Thank you in advance!