UA-AUG-AUMAT229 Lagrange's Theorem: Discovery guide

Topic 11 Lagrange's Theorem: Discovery guide

Definition 11.0.1. Cosets.

Given an element $g$ and a subgroup $H$ of a group $G,$ write $g H$ to mean the collection of all results of computing products

g h,

where $h$ is an element of $H .$ This collection is called the left coset of $H$ corresponding to $g .$

Similarly, the right coset $H g$ is the collection of all results of computing products

h g,

where $h$ is an element of $H .$

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Warning 11.0.2.

Unless $G$ is Abelian, in general left and right cosets $g H$ and $H g$ might be different collections.

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Note 11.0.3.

In additive notation, write $g + H$ for the left coset consisting of all sums $g + h .$ And similarly for right cosets.
The theory of right cosets is essentially the same as the theory of left cosets, so for now we will just concentrate on left cosets.

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Discovery 11.1.

Recall that

D_{6} = {e, r, r^{2}, r^{3}, r^{4}, r^{5}, s, r s, r^{2} s, r^{3} s, r^{4} s, r^{5} s},

with

\begin{aligned} (✶) & r^{6} & = e, & s^{2} & = e, & s r = r^{5} s, \end{aligned}

is the group of rotational symmetries of a thin hexagonal plate.

Let $H$ represent the cyclic subgroup $⟨ r^{2} ⟩ .$

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(a)

Write out the elements of $H,$ then compute the left coset $e H .$ What do you notice?

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(b)

Compute the left cosets $s H$ and $(r^{2} s) H,$ using the relations in (✶) to simplify each product computation result to one of the above-listed elements of $D_{6} .$ What do you notice? Why do you think this happened?

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(c)

Compute the left cosets $r^{2} H$ and $r^{4} H .$ What do you notice? Why do you think this happened?

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(d)

Compute the left coset $(r s) H$ and compare with $s H$ that you already computed in Task b. What do you notice? Why do you think this happened?

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(e)

Write out all four cosets $H,$ $s H,$ $r H,$ and $(r s) H$

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(i)

Is it possible to create a coset that is different from each of these four?

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(ii)

How many elements are in each of these cosets?

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(iii)

How many elements are in the subgroup $H ?$

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(iv)

How many cosets are there?

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(v)

How many elements are there in the group $G = D_{6} ?$

What is/are the relationship(s) between these numbers?

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Discovery 11.2.

Let's conjecture the general patterns of left cosets of a subgroup $H$ in a group $G$ from your discoveries in Discovery 11.1.

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(a)

Even if $g_{1} \neq g_{2},$ it could be that cosets $g_{1} H$ and $g_{2} H$ are the same collection of elements from $G .$ If you have calculated $g_{1} H,$ you can tell whether $g_{2} H$ will be the same coset without further calculations by checking .

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(b)

When $g_{1} H$ and $g_{2} H$ are not the same collection, what further relationships will these two cosets have to each other?

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(c)

Because the subgroup $H$ contains the identity element of $G,$ the coset $g H$ always contains .

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(d)

If $G$ contains $n$ elements, $H$ contains $m$ elements, and there are $c$ cosets, then the relationship between these three numbers is that .

In particular, the fraction

\frac{| G |}{| H |}

is always actually an .

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Discovery 11.3.

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(a)

Argue that in a finite group, the order of each element divides evenly into the order of the group.

Hint.

Consider Task 11.2.d for a cyclic subgroup $H = ⟨ x ⟩ .$

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(b)

Verify the statement of Task a for the example of $D_{6} .$

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Discovery 11.4.

Suppose $G$ is a finite group of order $n,$ $x$ is an element in $G$ of order $m,$ and there are $c$ different cosets of the cyclic subgroup $⟨ x ⟩$ in $G .$ Use the relationship between these three numbers from Task 11.2.d to determine what the result of $x^{n}$ must be.

Hint.

If you're not sure how Task 11.2.d applies here, consider the relationship between the order of an element and the order of the cyclic subgroup generated by that element.

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Warning 11.0.4. Avoid confusion.

Just to make sure Discovery 11.4 doesn't lead you to the wrong conclusions: a common mistake is to take a result like $x^{k} = e$ and conclude that $k$ must be equal to the order of $x .$ However, remember that the order of an element is the smallest positive exponent that takes the element to the identity, and any exponent that is a multiple of that order will also take the element to the identity.

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Discovery 11.5.

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(a)

Based on the statement of Task 11.3.a, what are the possibilities for the orders of the elements in the additive group $Z_{7} ?$ Which elements in that group have the various possible orders?

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(b)

Suppose $G$ is a group with order $p,$ where $p$ is a prime number. Argue that $G$ must be cyclic, and can be generated by any choice of non-identity element.

Hint.

Suppose $x$ is a non-identity element in $G .$ What are the possibilities for the order of $x ?$ How many elements are in the subgroup $⟨ x ⟩ ?$ How many elements are in $G ?$

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Discovery 11.6.

Recall that we can perform both operations of addition and multiplication in $Z_{n}$ via clock arithmetic, but the operation of multiplication does not make this collection into a group because some elements will not have a multiplicative inverse. However, as before we can create a multiplicative group $Z_{n}^{\times}$ (the textbook denotes this group by $R_{n}$ ) by throwing away those deficient elements that do not have a multiplicative inverse. In Discovery 3.5 we discovered that, in particular, we should discard the elements that share a common factor with $n .$

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(a)

Write out the elements of $Z_{16}^{\times} .$ How many are there?

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(b)

Based on Task 11.2.d, what are the possible sizes a subgroup of $Z_{16}^{\times}$ could have? For each possibility, give an example with that size.

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Discovery 11.7.

Contrary to the example of Task 11.6.b, just because a number divides evenly into the size of a finite group, it is not necessarily true that the group has a subgroup of that size.

Recall that $| S_{n} | = n!$ in general, so in particular $| S_{4} | = 24 .$ The subgroup of even permutations makes up half that number: $| A_{4} | = 12 .$

See if you can verify that, even though $6$ divides evenly into $12,$ the group $A_{4}$ has no subgroup of order $6 .$ (Don't worry if you get stuck, there is an explanation in the textbook.)