If a normal subgroup of a symmetric group contains a transposition. Prove that the normal subgroup is the entire symmetric group.

Question

Here is how I have done this problem. I would appreciate if someone can comment on my proof.

Let $N\triangleleft S_n$ and transposition $(a_i b_i) \in N$ where $a_i,b_i\in (1,2,....,n)$

Then $\forall \alpha \in S_n$

$\alpha^{-1}(a_i b_i)\alpha=N$

I can write $\alpha =(a_1a_2...a_k)(b_1b_2...b_t)....(c_1c_2...c_s)$ where the cycles are disjoint. Thus $\alpha^{-1} =(c_1c_2...c_s)^{-1}....(b_1b_2...b_t)^{-1}(a_1a_2...a_k)^{-1}$ Hence $\alpha^{-1} =(c_sc_{s-1}...c_1)....(b_tb_{t-1}...b_1)(a_ka_{k-1}...a_1)$

Thus

$\alpha^{-1}(a_i b_i)\alpha=N$ $\Rightarrow $

$(c_sc_{s-1}...c_1)...(b_tb_{t-1}...b_1)(a_ka_{k-1}...a_1)(a_i b_i)(a_1a_2...a_k)(b_1b_2...b_t)...(c_1c_2...c_s)= N$ $\Rightarrow $ $(a_{i-1} b_{i-1})=N$

Thus if $N$ contains a transposition $(a_i b_i)$ it implies it contains transposition $(a_{i-1} b_{i-1})$. Hence by induction we can deduce that $N$ contains all the transpositions. Since the smallest subgroup in $S_n$ containing all transpositions is $S_n$ itself hence we can conclude $N=S_n$.

I would appreciate if someone can comment on this proof. Thank you.

Answer 1

Observe that for any $\sigma\in S_n$ we have: $$\sigma^{-1}(a,b)\sigma=(\sigma(a),\sigma(b))$$ Hence if $N\unlhd S_n$ and $(a,b)\in N$ then $N$ contains all transpositions, but then $$N\geq\langle(a,b)|1\leq a<b\leq n\rangle=S_n$$

Answer 2

You have the right idea. The goal is to show the subgroup contains all transpositions. Suppose it contains $(ij)$ and we want to show it contains $(ab)$. Find a permutation $w$ with $w(i)=a$ and $w(j)=b$. Then $w(ij)w^{-1}=(ab)$. Hence the subgroup contains all transpositions and therefore must be the whole group.