# Intuitive idea on generators of $S_4$

What is the way to convince myself that $\left\langle(1,2),\ (1,2,3,4)\right\rangle=S_4$ but $\left\langle(1,3),\ (1,2,3,4)\right\rangle\ne S_4$?

Let $\sigma$ be any transposition and $\tau$ be any $p-$cycle, where $p$ is a prime. Then show that $S_p=\langle\sigma,\tau\rangle$.

• I don't understand you second question. Consider rephrasing. – bourbaki4481472 Dec 4 '13 at 0:33
• Try looking for an element that you cannot generate using $\langle(1,3)(1,2,3,4)\rangle$ in $S_4$. – bourbaki4481472 Dec 4 '13 at 0:40
• Note that a subset $S \subset S_4$ satisfies $\left\langle S\right\rangle=S_4$ iff $\left\langle S \right\rangle$ contains all transpositions in $S_4$ (since the transposition are a generating subset). Perhaps this will help with your intuition. – Omnomnomnom Dec 4 '13 at 0:45
• I observed that $(1,2,3)$ cannot be written as a product of the above two. How do I prove it? – Grobber Dec 4 '13 at 3:16

Besides to @Betty's points, there is another way for seeing why does this happen. We know that $S_4$ can have the following presentation:
$$S_4=\langle a,b\mid a^2=b^4=(ab)^3=1\rangle$$ Let's satisfy $a=(1,2),~~b=(1,2,3,4)$ in above relations. Indeed $a$ and $b$ can do that, but what will happen if we set $a=(1,3),~~b=(1,2,3,4)$? By this assumption, we see that $(ab)^3=(1,4)(2,3)$ and this happens cause of the points @Betty indicated them in detailed. Now if you are familiar to one of $D_8$'s presentation, then you'll have $$D_8=\langle a,b\mid a^2=b^4=(ab)^3=1\rangle,~~a=(1,3),~~b=(1,2,3,4)$$ instead wich is of order $8$.
• Our friend, $S_4$! +1 – amWhy Dec 7 '13 at 0:04
• Sir, if we have presentation given by $G=\langle a,b\mid a^2=b^4=(ab)^3=1\rangle$ then what we say about $G$, $G$ is isomorphic to $S_4$ or to $D_8$? – Akash Patalwanshi Feb 7 at 5:44