1.Let f ∈ S3 be a permutation which is not the identity. Prove that there exists g ∈ S3
such that fg ≠ gf.
2.Let n ≥ 3, and let f ∈ Sn be a permutation which is not the identity. Prove that
there exists g ∈ Sn such that fg ≠ gf.
Hey guys I am completely stuck in these two questions I'm really not sure where to start with solving it, can anyone please give me a hint? Any helps would be appreciated!
Update: Thanks everyone for giving me hint and helping me with that question. I have attempted to work on the first question and here is what I've got:
Since f ∈ S3 and it is not identity. This can conclude that there would be 5 permutations in f and 6 permutations for g as g ∈ S3.
We then test the permutations by multiple one f permutation to all g permutations. For example we take f = (321) and we multiple it with all g permutations available.
We would then be able to prove fg ≠ gf in most sets except the one where g is identity and the one that g has the same permutation as f.
Then, we can take this set to conclude that fg ≠ gf in f ∈ S3 and f is not identity and g ∈ S3 except when g is identity and the one that g has the same permutation as f.
Would this be correct at all?
Best Answer
Write $f$ as a bijective function from $\{1,2,...,n\}$ to itself. Since $f$ is not the identity, we have at least one $k \in \{1,2,...,n\}$ such that $f(k)\ne k$. Let $g \in S_n$ such that $g(k)=k$ and $g(f(k)) \ne f(k)$ (note that we need $n\ge 3$ here. Why?).
Now we have
$f(g(k))=f(k) \ne g(f(k))$
so $f$ and $g$ don't commute.