## Section6.2Lagrange'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 6.11, 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 6.12 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]\text{.} \end{equation*}

### Remark6.14.The 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 $$(1 \, 2 \, 3)$$ is in $$H\text{.}$$ Then $$(1 \, 2 \, 3)^{-1} = (1 \, 3 \, 2)$$ 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*} (1 \, 2 \, 4)(1 \, 2 \, 3)(1 \, 2 \, 4)^{-1} & = (1 \, 2 \, 4)(1 \, 2 \, 3)(1 \, 4 \, 2) = (2 \, 4 \, 3)\\ (2 \, 4 \, 3)(1 \, 2 \, 3)(2 \, 4 \, 3)^{-1} & = (2 \, 4 \, 3)(1 \, 2 \, 3)(2 \, 3 \, 4) = (1 \, 4 \, 2) \end{align*}

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

\begin{equation*} (1), (1 \, 2 \, 3), (1 \, 3 \, 2), (2 \, 4 \, 3), (2 \, 4 \, 3)^{-1} = (2 \, 3 \, 4), (1 \, 4 \, 2), (1 \, 4 \, 2)^{-1} = (1 \, 2 \, 4)\text{.} \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 )\text{.} \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\text{.} \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) )\text{.} \end{equation*}

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