How to prove fg doesn't equal to gf in permutation?

Question

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?

Answer 1

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.

Answer 2

Case $n=3$

This is best treated with disjoint cycle notation.

Consider $(12)$; then $(12)(13)=(132)$ and $(13)(12)=(123)$.

Consider $(123)$; then $(123)(12)=(13)$ and $(12)(123)=(23)$.

Up to a renaming of the elements, these are the only needed cases.

Case $n\ge3$.

Suppose $f$ is not the identity. By a renaming of the elements, we can assume that $f(1)=2$. Take the permutation $g$ that exchanges $2$ and $3$ and leaves everything else fixed. Then $$ g(f(1))=g(2)=3,\qquad f(g(1))=f(1)=2 $$ Therefore $gf\ne fg$.