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Prove/Disprove: Find $a\in S_4$ that is not a power of $b$ such that $ab=ba$ while $b=(123)$

What I did:

I think this statement is false.

So we need to find $aba^{-1}=b$, we know that $id\in S_4$ will surely $id*b*id^{-1}=b$ but we get $id^k=b$, I wonder if there are more elements such that $ab=ba$, how to find them?

Any help will be appreciated.

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    @pjs36 I think you're right, I deleted my comment. – 2017-01-28

1 Answers 1

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If $a$ commutes with $b$, let's look where $ab=ba$ sends 4. We see that $ba(4)=ab(4)=a(4)$, thus $a(4)$ is a stable point of $b$, thus $a(4)=4$. So we reduced the problem to the case of $S_3$, but for $S_3$ it is straitforward to check that any transposition does not commute with $(123)$, and all others are powers of $b$.

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    So I need to check for all possible elements in $S_3$? Meaning to check for 6 different elements to check commute and if so to check for each element if it a power of $b$ – 2017-01-28
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    Nope, in $S_3$ you have 6 elements: powers of $b$: $id,b,b^2$ and three transpositions. Powers of $b$ obviously commute with $b$, so nothing to check for them. But if you checked that (123) and (12) do not commute, then obviously (123)=(231) and (23) do not commute, because we just renamed elements, similarly, (312) and (31) do not commute. – 2017-01-28