# Proving that $H$ subgroup is equal to $S_n$

Let $H$ a subgroup of $S_n$ such that for all $a,b\in \{1,2,...,n\}$ there exist $\tau\in H$ such that $\tau(a)=b$. Show that if $H$ contains a transposition and a $n-1$ cycle then $H=S_n$.

• If the first fact is true then I see no need for the second part. Since every element in $S_n$ can be rewritten as the product of transpositions and the fact that all transpositions in $S_n$ are contained in $H$($\exists\tau\in H$ such that $\tau(a)=b$). Would be enough to prove the statement. (Okay ignore this I see that the way you used $\tau$ does not imply that it is a transposition ) – user160110 Oct 25 '15 at 21:39
• @user160110 : The fact that there exist an element of $\tau \in H$ such that $\tau(a) = b$ does not imply that $\tau$ is a transposition. For example, the subgroup of permutations of the form $\tau_a : k \mapsto k + a \mod n$ (where ${1,\ldots, n}$ is identified with $\mathbb{Z}/n\mathbb{Z}$) has that property (take $\tau_{b-a}$) but does not contain any transpositions (except if $n = 2$ of course). – Joel Cohen Oct 25 '15 at 21:49
• @JoelCohen I know this, but the symbol $\tau$ is often used for transpostions where as symbols like $\alpha, \beta,$ and $\phi$ might be used for general permutaions. I just misread the second part. – user160110 Oct 25 '15 at 21:52
• @user160110 $\tau$ not is necesarilly a transposition – Roiner Segura Cubero Oct 26 '15 at 0:38

Okay since you have a transposition (lets call it $\tau=(i,j)$ ) in H and for any $a,b\in\{1,...,n\}$ their exist a permutation in $\beta_{a,b}\in H$ such that $\beta_{a,b}(a)=b$, it follows that for any $a\in \{1,...,n\}$ we have that $\beta_{i,a}\tau \beta_{i,a}^{-1}=(a,j)$. In this way we generate all transpositions and thus all permutations.
• @RoinerSeguraCubero Let $a,b\in \{1,...,n\}$ then $\beta_{j,b}[\beta_{i,a}(i,j)\beta_{i,b}^{-1}]\beta_{j,a}^{-1}=(a,b)$ – user160110 Oct 26 '15 at 1:20
• $\beta$ might move $j$ too, so you only get that $(a,k(a))\in H$ for all $a$. – user138530 Oct 26 '15 at 2:49
• But you can finish up as follows: do what you did for the $a$ not in the $n-1$ cycle, and then move $k(a)$ around at will by conjugating with the $n-1$ cycle. This shows that $(a,b)\in H$ for some fixed $a$ and all $b$, and these generate $S_n$. – user138530 Oct 26 '15 at 2:53