Abstract Algebra: Theory and Applications

We first show that the map $\varphi$ is one-to-one. Suppose that $\varphi (h_1)=\varphi (h_2)$ for elements $h_1,h_2\in H\text{.}$ We must show that $h_1=h_2\text{,}$ but $\varphi (h_1)=g{h}_1$ and $\varphi (h_2)=g{h}_2\text{.}$ So $g{h}_1=g{h}_2\text{,}$ and by left cancellation $h_1=h_2\text{.}$ To show that $\varphi$ is onto is easy. By definition every element of $gH$ is of the form $gh$ for some $h\in H$ and $\varphi (h)=gh\text{.}$

The group $G$ is partitioned into $[G:H]$ distinct left cosets. Each left coset has $|H|$ elements; therefore, $|G|=[G:H]|H|\text{.}$

Let $g$ be in $G$ such that $g\ne e\text{.}$ Then by Corollary 6.11, the order of $g$ must divide the order of the group. Since $|⟨g⟩|>1\text{,}$ it must be $p\text{.}$ Hence, $g$ generates $G\text{.}$

Observe that

Since $[A_4:H]=2\text{,}$ there are only two cosets of $H$ in $A_4\text{.}$ Inasmuch as one of the cosets is $H$ itself, right and left cosets must coincide; therefore, $gH=Hg$ or $gH{g}^{-1}=H$ for every $g\in A_4\text{.}$ Since there are eight $3$-cycles in $A_4\text{,}$ at least one $3$-cycle must be in $H\text{.}$ Without loss of generality, assume that $(123)$ is in $H\text{.}$ Then $(123{)}^{-1}=(132)$ must also be in $H\text{.}$ Since $gh{g}^{-1}\in H$ for all $g\in A_4$ and all $h\in H$ and

we can conclude that $H$ must have at least seven elements

Therefore, $A_4$ has no subgroup of order $6\text{.}$

Suppose that

Define $\sigma$ to be the permutation

Then $\mu =\sigma \tau {\sigma }^{-1}\text{.}$

Conversely, suppose that $\tau =(a_1,a_2,\dots ,a_k)$ is a $k$-cycle and $\sigma \in S_n\text{.}$ If $\sigma (a_i)=b$ and $\sigma (a_{(imodk)+1})=b^{\prime }\text{,}$ then $\mu (b)=b^{\prime }\text{.}$ Hence,

Since $\sigma$ is one-to-one and onto, $\mu$ is a cycle of the same length as $\tau \text{.}$