# 6.2: Lagrange's Theorem

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## Proposition $$6.9$$

Let $$H$$ be a subgroup of $$G$$ with $$g \in G$$ and define a map $$\phi:H \rightarrow gH$$ by $$\phi(h) = gh\text{.}$$ The map $$\phi$$ is bijective; hence, the number of elements in $$H$$ is the same as the number of elements in $$gH\text{.}$$

Proof

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 $$\phi(h) = gh\text{.}$$

## Theorem $$6.10$$. Lagrange

Let $$G$$ be a finite group and let $$H$$ be a subgroup of $$G\text{.}$$ Then $$|G|/|H| = [G : H]$$ is the number of distinct left cosets of $$H$$ in $$G\text{.}$$ In particular, the number of elements in $$H$$ must divide the number of elements in $$G\text{.}$$

Proof

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

Suppose that $$G$$ is a finite group and $$g \in G\text{.}$$ Then the order of $$g$$ must divide the number of elements in $$G\text{.}\ ## Corollary \(6.12$$

Let $$|G| = p$$ with $$p$$ a prime number. Then $$G$$ is cyclic and any $$g \in G$$ such that $$g \neq e$$ is a generator.

Proof

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 $${\mathbb Z}_p\text{.}$$

## Corollary $$6.13$$

Let $$H$$ and $$K$$ be subgroups of a finite group $$G$$ such that $$G \supset H \supset K\text{.}$$ Then

$[G:K] = [G:H][H:K]\text{.} \nonumber$

Proof

Observe that

$[G:K] = \frac{|G|}{|K|} = \frac{|G|}{|H|} \cdot \frac{|H|}{|K|} = [G:H][H:K]\text{.} \nonumber$

## Remark $$6.14$$. 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.

## Proposition $$6.15$$

The group $$A_4$$ has no subgroup of order $$6\text{.}$$

Proof

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

$(1), (123), (132), (243), (243)^{-1} = (234), (142), (142)^{-1} = (124)\text{.} \nonumber$

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

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

## Theorem $$6.16$$

Two cycles $$\tau$$ and $$\mu$$ in $$S_n$$ have the same length if and only if there exists a $$\sigma \in S_n$$ such that $$\mu = \sigma \tau \sigma^{-1}\text{.}$$

Proof

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,

$\mu = ( \sigma(a_1), \sigma(a_2), \ldots, \sigma(a_k) )\text{.} \nonumber$

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

This page titled 6.2: Lagrange's Theorem is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by Thomas W. Judson (Abstract Algebra: Theory and Applications) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.