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Question: Let f ∈ S3 be a permutation which is not the identity. Prove that there exists g ∈ S3 such that fg ≠ gf.

My attempt:

First list out all the possible permutations of f (which there are 5 of them as it is not identity); then list out all possible permutations of g (which there are 6 of them). Then I try each set of multiplication of fg and gf to check they are not equal in order to prove the hypothesis.

However, I do feel like I'm using the wrong method to prove this hypothesis as each permutation of f could multiple with 6 different permutations of g; It would take forever to complete every sets of multiplication by this method. May I ask is there an easier way to solve this question? I feel like I'm missing something out...

sasaki085
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  • yah that was an example that I was thinking about. But can the others contradict this example by saying something like "you are only showing one possible set of multiplication that would not prove that all sets of f permutations and g permutations multiplications would result in ≠ ? – sasaki085 Nov 18 '20 at 07:37
  • Actually show that $(1, 2)$ doesn't commute with any member of $S_3$ except identity and itself. Then you prove indirectly that for every non-identity element there exist another element with which it doesn't commute. Can you see why? – PNDas Nov 18 '20 at 07:49

1 Answers1

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Actually show that $(1\, 2)$ doesn't commute with any member of $S_3$ except identity and itself. Then you prove indirectly that for every non-identity element there exist another element with which it doesn't commute. Can you see why?

Suppose $f \in S_3$ and $f\neq e$ and $f\neq (1\, 2)$ then take $g=(1\, 2)$.

If $f = (1\, 2)$ then take any other non-identity element as $g$ as you have shown $(1\, 2)$ doesn't commute with any member of $S_3$ except identity and itself so you are done.

EDITED ON 21-11-2020: Another answer which is a proof.

Theorem: $\forall n\geq 3, Z(S_n)={e}$, where $Z(G)$ is center of $G$ and $e$ is identity.

That means identity is the only element that commutes with every elemnt of $S_n\forall n\geq 3$.(From now on whenever I say $S_n$, just assume $n\geq 3$).

In other words for every non-identity element $f\in S_n$ there exists another non-identity element $g\in S_n$ s.t. $fg\neq gf$.

Proof:-

First observe that to prove $fg=gf$, we need to show that $fg(x)=gf(x)\forall x\in \{1,2,3,\cdots ,n\}$.

But to prove $fg\neq gf$, we need to show that $\exists x\in \{1,2,3,\cdots ,n\}\text{s.t.} fg(x)\neq gf(x)$.

Let $a,b,c\in \{1,2,3,\cdots ,n\}$.

Let an arbitrary $f\neq e\in S_n$ s.t. $f(a)=b$, Now choose $g=(b,c)$ where $c\neq f(b)$. Observe that this is possible only because $n\neq 3$.

Let me give you an example suppose in $S_3$, let $f(1)=3$ now $f(3)=1\text{ or }2$ say $1$, >then choose $g=(3,2)$.

Now, observe $$fg(a)=f(g(a))=f(a)=b\\gf(a)=g(f(a))=g(b)=c$$

Hence $fg\neq gf.$. So for every non-identity element $f\in S_n$ there exist another non-identity element $g\in S_n$ s.t. $fg\neq gf$.$\qquad \square$

PNDas
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  • I kinda understand what are you saying. Um.. I know this may be a stupid question but how did you figure out to use (1,2) among all possible sets? – sasaki085 Nov 18 '20 at 07:56
  • What I said is true for any other element of order 2 that is $(1,2),(2,3),(1,3)$.But you can't take $(1,2,3)$ or $(1,3,2)$ as they are inverses of each other so they commute with each other. So if $f=(1,2,3)$ then you can't take $g=(1,3,2)$. – PNDas Nov 18 '20 at 07:59
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    @sasaki085, I edited my answer with a proof actually this is available on this site but I came across this problem in GALLIAN today. So I decided to include the proof here. – PNDas Nov 21 '20 at 15:39