Exercise 3.5.16 (Generators of $A_5$)

Proof. Let $x$ and $y$ be distinct $3$-cycles in $S_5$ with $x\ne y^{-1}$.

(a)

Up to conjugation, we may assume that $x$ and $y$ fix the element $5$. Then by Exercise 15, $x,y$ generate the subgroup of even permutations of $S_5$ which fix $5$, so $$⟨x,y⟩\simeq A_4.$$

(b)

First we show that $$\begin{array}{lll}\hfill A_5=⟨(123),(145)⟩.& & \hfill \text{(1)}\end{array}$$

Put $H=⟨r,s⟩$, where $r=(123),s=(145)$. Since $r$ and $s$ are even, $H\le A_5$. Note that

$$\mathit{𝑠𝑟}=(145)(123)=(12345),$$

so $H$ contain an element of order $5$.

Moreover

$$\begin{array}{llll}\hfill s{r}^{-1}{s}^{-1}{r}^{-1}& =(s(132)s^{-1})(132)& \hfill & \\ \hfill & =(432)(132)& \hfill & \\ \hfill & =(12)(34).& \hfill & \end{array}$$

So $H$ contains the element $(12)(34)$ of order 2, the element $(123)$ of order 3, and the element $(12345)$ of order $5$. By Lagrange’s Theorem, if $n=|H|$, then $n\mid 60=|A_5|$, and

$$2\mid n,3\mid n,5\mid n,$$

therefore $30\mid n$ and $n\mid 60$, so

$$|H|=30\text{or}|H|=60.$$

It remains to prove that $A_5$ has no subgroup of order $30$.

(If we know that $A_5$ is simple, this is obvious: if $|H|=30$ then $|A_5:H|=2$, therefore $H⊴A_5$: this is impossible since $A_5$ is simple. We give a proof which don’t use this fact.)

Put $G=A_5$. Assume for the sake of contradiction that $|H|=30$, so that $|G:H|=2$. Then $H⊴G$ and $|G∕H|=2$. Every element $\bar{g}\in G∕H$ satisfies ${\bar{g}}^2=\bar{1}$ by Lagrange’s Theorem, so for all $g\in G$,

$$\begin{array}{lll}\hfill g^2\in H(g\in G).& & \hfill \text{(2)}\end{array}$$

Moreover,

$$\begin{array}{llll}\hfill (abc)& ={(acb)}^2,& \hfill & \\ \hfill (abcde)& ={(adbec)}^2,& \hfill & \end{array}$$

so every $3$-cycle and every $5$ cycle is a square in $G$, so is in $H$. By Exercise 1.3.16, there are $20$ ( $=\frac{5\times 4\times 3}{3}$) $3$-cycles and $24$( $=\frac{5\times 4\times 3\times 2\times 1}{5})$ $5$-cycles, so there are at least $44=20+24$ distinct squares in $H$. By (2), $|H|\ge 44$, in contradiction with $|H|=30$. This shows that $A_5$ has no subgroup of order $30$, hence $|H|=60$, so $A_5=H$:

$$\begin{array}{lll}\hfill A_5=⟨(123),(145)⟩.& & \hfill \end{array}$$

Now assume that $x$ and $y$ do not fix a common element of $[[1,5]]$. Set $H=⟨x,y⟩$. Since $x$ and $y$ are $3$-cycles, they fix $2$ elements, so $x$ fixes $a,b$ and $y$ fixes $c,d$, where $\{a,b\}\cap \{c,d\}=\varnothing$. It remains a unique element $e$ in the support of $x$ and $y$, so

$$\begin{array}{lll}\hfill x=(ecd)\text{or}x=(edc)=(ecd),& & \hfill \\ \hfill y=(eab)\text{or}y=(eba)={(eab)}^{-1}.& & \hfill \end{array}$$

In every case, $H$ contains $(ecd)$ and $(eab)$, which are conjugate to $(123)$ and $(145)$: if

$$\alpha =\left(\begin{array}{ccccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill & \hfill 5\hfill \\ \hfill e\hfill & \hfill c\hfill & \hfill d\hfill & \hfill a\hfill & \hfill b\hfill \end{array}\right),$$

then

$$\begin{array}{llll}\hfill x& =(ecd)=\alpha (123){\alpha }^{-1},& \hfill & \\ \hfill y& =(eab)=\alpha (145){\alpha }^{-1}.& \hfill & \end{array}$$

Then the argument given in Exercise 5 (or in Exercise 14) shows that (1) implies

$$A_4=⟨x,y⟩.$$