Exercise 3.5.14 ($A_4 = \langle \sigma, \tau \rangle $, where $|\sigma| = 3$, $|\tau| = 2$.)

Proof. Let $H=⟨\sigma ,\tau ⟩$, where $|\sigma |=2$ and $|\tau |=2$.

Let $n=|H|$. By Lagrange’s Theorem, $n\mid 12=|A_4|$, and

Therefore $6\mid n$, and $n\mid 12$, so

But we know that $A_4$ has no subgroup of order $6$ (see Exercise 6). Therefore $n=12$, so $|H|=|A_4|$, where $H\le A_4$. This proves $H=A_4$:

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Proof. First we show that

Set $H=⟨(123),(12)(34)⟩$. Then $(132)={(123)}^{-1}\in H$, so

(see the Lemma of Exercise 3.5.4).

Therefore

The lattice of subgroups of $A_4$ given in the text shows that the only subgroup containing $⟨(12)(34),(13)(24)⟩$ and $⟨(123)⟩$ is $A_4$. So (1) is proven.

Now take any element $\sigma \in A_4$ of order $3$, and any element $\tau \in A_4$ of order $2$. Then $\sigma =(abc)$, where $\{a,b,c\}\subset \{1,2,3,4\}$, and $\tau$ is the product of two disjoint transpositions, hence the support of one of them is included in $\{a,b,c\}$. So we can write without loss of generality $\tau =(ab)(cd)$, where $a,b,c,d$ are distinct, so $\{a,b,c,d\}=\{1,2,3,4\}$.

Consider le permutation

By the Lemma of Exercise 3.5.3,

Let $f$ be the map defined by ${\gamma }_{\alpha }(\delta )=\alpha \delta {\alpha }^{-1}$ for every permutation $\delta \in A_4$ (then $\alpha \delta {\alpha }^{-1}$ is even). Then $f$ is an automorphism of $A_4$ (but not necessarily an inner automorphism).

Let $\xi$ by any permutation in $S_n$. Since $f$ is an automorphism, there is some permutation $\zeta \in A_4$ such that $\xi =\mathit{𝛼𝜁}{\alpha }^{-1}=f(\zeta )$. Since $A_4=⟨(123),(12)(34)⟩$, by Proposition 9,

where ${𝜀}_i\in \{-1,1\}$ and ${\sigma }_i\in \{(123),(12)(34)\}$ for $i\in [[1,k]]$. Therefore

where ${\tau }_i=f({\sigma }_i)\in \{\sigma ,\tau \}$. This shows that

$A_4$ is generated by $\sigma ,\tau$, where $\tau$ is any element of order $2$ and $\sigma$ is any element of order $3$. □