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## Section7.4Lagrange's Theorem

We first show that the map $\phi$ is one-to-one. Suppose that $\phi(h_1) = \phi(h_2)$ for elements $h_1, h_2 \in H\text{.}$ We must show that $h_1 = h_2\text{,}$ but $\phi(h_1) = gh_1$ and $\phi(h_2) = gh_2\text{.}$ So $gh_1 = gh_2\text{,}$ and by left cancellation $h_1= h_2\text{.}$ To show that $\phi$ is onto is easy. By definition every element of $gH$ is of the form $gh$ for some $h \in H$ and $\phi(h) = gh\text{.}$

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 $g \neq e\text{.}$ Then by Corollary 7.11, the order of $g$ must divide the order of the group. Since $|\langle g \rangle| \gt 1\text{,}$ it must be $p\text{.}$ Hence, $g$ generates $G\text{.}$

Corollary 7.12 suggests that groups of prime order $p$ must somehow look like ${\mathbb Z}_p\text{.}$

Observe that

\begin{equation*} [G:K] = \frac{|G|}{|K|} = \frac{|G|}{|H|} \cdot \frac{|H|}{|K|} = [G:H][H:K]. \end{equation*}
###### Remark7.14The converse of Lagrange's Theorem is false

The group $A_4$ has order $12\text{;}$ however, it can be shown that it does not possess a subgroup of order $6\text{.}$ According to Lagrange's Theorem, subgroups of a group of order $12$ can have orders of either $1\text{,}$ $2\text{,}$ $3\text{,}$ $4\text{,}$ or $6\text{.}$ However, we are not guaranteed that subgroups of every possible order exist. To prove that $A_4$ has no subgroup of order $6\text{,}$ we will assume that it does have such a subgroup $H$ and show that a contradiction must occur. Since $A_4$ contains eight $3$-cycles, we know that $H$ must contain a $3$-cycle. We will show that if $H$ contains one $3$-cycle, then it must contain more than $6$ elements.

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 $g H g^{-1} = H$ 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 $g h g^{-1} \in H$ for all $g \in A_4$ and all $h \in H$ and

\begin{align*} (124)(123)(124)^{-1} & = (124)(123)(142) = (243)\\ (243)(123)(243)^{-1} & = (243)(123)(234) = (142) \end{align*}

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

\begin{equation*} (1), (123), (132), (243), (243)^{-1} = (234), (142), (142)^{-1} = (124). \end{equation*}

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

In fact, we can say more about when two cycles have the same length.

Suppose that

\begin{align*} \tau & = (a_1, a_2, \ldots, a_k )\\ \mu & = (b_1, b_2, \ldots, b_k ). \end{align*}

Define $\sigma$ to be the permutation

\begin{align*} \sigma( a_1 ) & = b_1\\ \sigma( a_2 ) & = b_2\\ & \vdots\\ \sigma( a_k ) & = b_k. \end{align*}

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

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

\begin{equation*} \mu = ( \sigma(a_1), \sigma(a_2), \ldots, \sigma(a_k) ). \end{equation*}

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