AATA Lagrange’s Theorem

Section 6.2 Lagrange’s Theorem

Proposition 6.9.

Let

H

be a subgroup of

G

with

g \in G

and define a map

ϕ : H \to g H

ϕ (h) = g h .

The map

ϕ

is bijective; hence, the number of elements in

H

is the same as the number of elements in

g H .

Proof.

We first show that the map

ϕ

is one-to-one. Suppose that

ϕ (h_{1}) = ϕ (h_{2})

for elements

h_{1}, h_{2} \in H .

We must show that

h_{1} = h_{2},

but

ϕ (h_{1}) = g h_{1}

and

ϕ (h_{2}) = g h_{2} .

g h_{1} = g h_{2},

and by left cancellation

h_{1} = h_{2} .

To show that

ϕ

is onto is easy. By definition every element of

g H

is of the form

g h

for some

h \in H

and

ϕ (h) = g h .

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Theorem 6.10. Lagrange.

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Let

G

be a finite group and let

H

be a subgroup of

G .

Then

| G | / | H | = [G : H]

is the number of distinct left cosets of

H

G .

In particular, the number of elements in

H

must divide the number of elements in

G .

Proof.

The group

G

is partitioned into

[G : H]

distinct left cosets. Each left coset has

| H |

elements; therefore,

| G | = [G : H] | H | .

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Corollary 6.11.

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Suppose that

G

is a finite group and

g \in G .

Then the order of

g

must divide the number of elements in

G .

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Corollary 6.12.

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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 .

Then by Corollary 6.11, the order of

g

must divide the order of the group. Since

| ⟨ g ⟩ | > 1,

it must be

p .

Hence,

g

generates

G .

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Corollary 6.12 suggests that groups of prime order

p

must somehow look like

Z_{p} .

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Corollary 6.13.

🔗

Let

H

and

K

be subgroups of a finite group

G

such that

G \supset H \supset K .

Then

[G : K] = [G : H] [H : K] .

Proof.

Observe that

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

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Remark 6.14. The converse of Lagrange’s Theorem is false.

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The group

A_{4}

has order

12;

however, it can be shown that it does not possess a subgroup of order

6 .

According to Lagrange’s Theorem, subgroups of a group of order

12

can have orders of either

1,

2,

3,

4,

6 .

However, we are not guaranteed that subgroups of every possible order exist. To prove that

A_{4}

has no subgroup of order

6,

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.

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Proposition 6.15.

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The group

A_{4}

has no subgroup of order

6 .

Proof.

Since

[A_{4} : H] = 2,

there are only two cosets of

H

A_{4} .

Inasmuch as one of the cosets is

H

itself, right and left cosets must coincide; therefore,

g H = H g

g H g^{- 1} = H

for every

g \in A_{4} .

Since there are eight

3

-cycles in

A_{4},

at least one

3

-cycle must be in

H .

Without loss of generality, assume that

(1 2 3)

is in

H .

Then

(1 2 3)^{- 1} = (1 3 2)

must also be in

H .

Since

g h g^{- 1} \in H

for all

g \in A_{4}

and all

h \in H

and

\begin{aligned} (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{aligned}

we can conclude that

H

must have at least seven elements

(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) .

Therefore,

A_{4}

has no subgroup of order

6 .

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

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Theorem 6.16.

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Two cycles

τ

and

μ

S_{n}

have the same length if and only if there exists a

σ \in S_{n}

such that

μ = σ τ σ^{- 1} .

Proof.

Suppose that

\begin{aligned} τ & = (a_{1}, a_{2}, \dots, a_{k}) \\ μ & = (b_{1}, b_{2}, \dots, b_{k}) . \end{aligned}

Define

σ

to be the permutation

\begin{aligned} σ (a_{1}) & = b_{1} \\ σ (a_{2}) & = b_{2} \\ ⋮ \\ σ (a_{k}) & = b_{k} . \end{aligned}

Then

μ = σ τ σ^{- 1} .

Conversely, suppose that

τ = (a_{1}, a_{2}, \dots, a_{k})

is a

k

-cycle and

σ \in S_{n} .

σ (a_{i}) = b

and

σ (a_{(i mod k) + 1}) = b^{'},

then

μ (b) = b^{'} .

Hence,

μ = (σ (a_{1}), σ (a_{2}), \dots, σ (a_{k})) .

Since

σ

is one-to-one and onto,

μ

is a cycle of the same length as

τ .

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