# Show that multiplication in $S_n$ is noncommutative $\forall n \geq 3$

group-theoryinductionpermutationssolution-verificationsymmetric-groups

The hint hat was given for this problem is 'HINT: Find a pair of 2-cycles that do not commute'

So, starting off I interpret this question to mean prove this by induction (given that it's a statement of proof about a proposition indexed by the natural numbers) so I will start with a base case.

Base Case: $$n = 3$$

The group $$S_3$$ has the elements $$(), (1 2), (1 3), (2 3), (1 2 3), (1 3 2)$$ so looking at the pair of transpositions $$(1 2)$$ and $$(2 3)$$ we see that: $$(1 2)(2 3) = (1 2 3) \neq (1 3 2) = (2 3)(1 2)$$

Inductive Hypothesis:

Assume that multiplication in $$S_r$$ is noncommutative $$\forall r \in \mathbb{N}$$ such that $$4 \leq r \leq k$$ for some $$k \geq 3$$

Inductive Step:

We want to show that multiplication is noncommutative in $$S_{k+1}$$. Note that any $$k$$-Cycle can be written as a product of $$(k-1)$$ $$2$$-cycles. So we can say that any $$(k+1)$$-cycle (i.e., any element of $$S_{k+1}$$) can be written as a product of $$(k)$$ $$2$$-cycles, but as we've shown transpositions are not commutative in general. Thus multiplication in $$S_{k+1}$$ is noncommutative.

I just want to make sure that

1. I was correct in that we could use a proof by induction to prove this statement
2. My setup for each step of the proof was right
3. The logic in the inductive step was right – I'm not sure if by showing that 2 transpositions in $$S_3$$ are noncommutative would be sufficient for saying that any transpositions in $$S_{k+1}$$ would also be noncommutative.

Thanks for any help!

Since $$(12)(23)\neq(23)(12)$$, it suffices to embed these elements into $$S_n$$ for $$n\ge 4$$ by, in essence, noticing that you can multiplying each of $$(12),(23)$$ by $$(i)$$ for $$i\in\{1,\dots, n\}$$; more concretely: define $$\varphi: S_3\to S_n$$ by
$$\varphi_\sigma(i) =\begin{cases} \sigma(i) &: i\in\{1,2,3\}\\ i &: i\in\{4,\dots, n\} \end{cases}$$
for each $$\sigma\in S_3$$. It is routine to check that $$\varphi$$ is an injective homomorphism.