# Inner Automorphism using Transpositions

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?

• Ah good point! My "disjoint and not equal" part is probably just incorrect. I was thinking of "if two cycles are disjoint then they commute," but I incorrectly extended that idea by replacing "cycles" with "permutations in general", and by making it an "iff" statement, when it isn't. So then it is possibly accurate to say that if $\tau$ and $\sigma$ do not commute, then $\tau \sigma \tau \neq \sigma$. But I need a better justification for why there exists $\tau$ that does not commute with $\sigma$? Or is that still not quite right? – 1Teaches2Learn Apr 2 at 3:28
• I agree with your comment, that if you can find a $\tau$ that does not commute with $\sigma$, then you are done. Do you know that $S_n$ has trivial center when $n \geq 3$? And do you know that $S_n$ is generated by transpositions? If you know both those things, you could try using proof by contradiction: Assume that $\sigma$ commutes with every transposition... – CJD Apr 2 at 3:33
• Great points. For this class, we do not "know" that $S_n$ has trivial center when $n \geq 3$ yet. By that I mean that it isn't something we've shown in class yet, so it wouldn't have been "fair game" to just state that on this test. (And in fact, the next part of the problem actually asked, "What can you conclude about the center of $S_n$ based on your answer to the preceding question?") HOWEVER, your point about $S_n$ being generated by transpositions is extremely interesting, and I will think more about that. I think that might be the missing piece. Thanks so much! – 1Teaches2Learn Apr 2 at 3:41