Math 417 – Tenth Day, Lecture notes of Algebra

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Math 417 – Tenth Day

Bruce Reznick University of Illinois at Urbana-Champaign

September 16, 2020

Notation that is standard and in the book: if H is a subgroup of G , we often write H ≤ G or G ≥ H, with H < G and G > H meaning that H is a subgroup of G and H 6 = G.

I’d like to return to cosets, in general. There will be examples later. Once again, recall the definition: Suppose H is a subgroup of G and g ∈ G (not necessarily in H). The left coset gH and the right coset Hg are defined by

gH = {gh | h ∈ H}, Hg = {hg | h ∈ H}.

Before we prove some important results about them, I’d like to give an example involving infinite groups that shows we’ve already been working with cosets.

Consider the group (Z, +); that is, the integers under addition. We have already seen that this is a group: 0 is the identity −n is the inverse of n ∈ Z and ordinary addition is associative. Suppose d ∈ N, so d ≥ 1.

LEMMA If G ≥ H and H is a finite group, then for every g ∈ G , we have |H| = |gH| = |Hg |; that is, every left and right coset has the same number of elements as H.

PROOF Suppose H = {h 1 ,... , hm}. Then gH = {gh 1 ,... , ghm}, so the statement is true provided hi 6 = hj =⇒ ghi 6 = ghj.

But the contrapositive of that statement is ghi = ghj =⇒ hi = hj , which we know to be true upon multiplication of both sides by g −^1. The same argument works for Hg.

The next result is sort of surprising and really important.

THEOREM If G ≥ H and xH and yH are two left cosets, then either xH = yH as sets, or xH and yH are disjoint; that is, xH ∩ yH = ∅. Furthermore, xH = yH if and only if x = yh, y = xh′ for some h, h′^ ∈ H. The same thing holds for right cosets, with the last condition changed to x = hy , y = h′x.

PROOF. Suppose z ∈ xH ∩ yH, so the intersection is not empty. By definition, this means that there exist h 1 , h 2 ∈ H so that

z = xh 1 = yh 2 =⇒ x = yh 2 h− 1 1.

Since H is a subgroup, h 2 h− 1 1 ∈ H. Suppose now that u ∈ xH. Then for some h ∈ H,

u = xh = (yh 2 h− 1 1 )h = y (h 2 h− 1 1 h)

and, again, H is a subgroup, so h 2 h− 1 1 h ∈ H. This implies that u ∈ yH. Thus xH ⊆ yH.

But also y = xh 1 h− 2 1 , so an almost identical argument shows that yH ⊆ xH, and so xH = yH.

We have seen that if x = yh or y = xh′, then xH = yH. The converse is easier: if xH = yH, then x ∈ xH =⇒ x ∈ yH =⇒ x = yh, y = xh−^1 for some h ∈ H.

The identical argument (up to the order in which you write elements) applies to right cosets.

We’ve seen examples of this earlier. If G = C 6 = 〈a〉, a^6 = e, then as we have already seen, G has four subgroups: H 1 = {e}, H 2 = {e, a^3 }, H 3 = {e, a^2 , a^4 } and G itself, and we have seen that

[G : H 1 ]L = [G : H 1 ]R = 6, [G : H 2 ]L = [G : H 2 ]R = 3 [G : H 3 ]L = [G : H 3 ]R = 2, [G : G ]L = [G : G ]R = 1.

The pattern isn’t an accident. The following was the first major theorem in the subject.

LAGRANGE’S THEOREM If G is a finite group and G ≥ H, then

[G : H]L = [G : H]R =

|G |

|H|

In particular, |H| is a divisor of |G |.

We have already seen this in the special case of the cyclic group.

PROOF. Suppose that |G | = n and H = {h 1 ,... , hm} and [G : H]L = r. Write down G as a union of the distinct cosets of H: a 1 H,... , ar H. a 1 H = {a 1 h 1 ,... a 1 hm} a 2 H = {a 2 h 1 ,... a 2 hm} · · · ar H = {ar h 1 ,... ar hm}.

Now count elements.

G =

⋃^ r

k=

ak H = {a 1 h 1 ,... a 1 hm, a 2 h 1 ,... a 2 hm... , ar h 1 ,... ar hm}

There are n elements in G and r · m = [G : H]L · |H| elements on the right hand side, so |G | = [G : H]L · |H|. The exact same argument would work with right cosets as well, so |G | = [G : H]R · |H|, hence [G : H]L = [G : H]R and we get the desired formula..

COROLLARY If G > H, and [G : H] = 2, then the cosets of H (left or right) are H and G \ H, the elements of G which are not in H.

PROOF The coset associated with e is eH = He = H, so we may take one of the left (or right) cosets to be H, so

G = H ∪ aH; G = H ∪ Hb

for some a, b ∈ G. But these are disjoint unions, so aH (and Hb) consist of the elements in G which are not in H.

Well this might seem very abstract, so let’s look at the cosets of the subgroups of S 3. l’ll cut and paste the multiplication table again.

ρ 0 =

= (1)(2)(3), ρ 1 =

ρ 2 =

= (132), μ 1 =

μ 2 =

= (13)(2), μ 3 =

S 3 ρ 0 ρ 1 ρ 2 μ 1 μ 2 μ 3 ρ 0 ρ 0 ρ 1 ρ 2 μ 1 μ 2 μ 3 ρ 1 ρ 1 ρ 2 ρ 0 μ 3 μ 1 μ 2 ρ 2 ρ 2 ρ 0 ρ 1 μ 2 μ 3 μ 1 μ 1 μ 1 μ 2 μ 3 ρ 0 ρ 1 ρ 2 μ 2 μ 2 μ 3 μ 1 ρ 2 ρ 0 ρ 1 μ 3 μ 3 μ 1 μ 2 ρ 1 ρ 2 ρ 0

We have already seen that the subgroups of S 3 are the trivial ones ({ρ 0 }, S 3 ), plus {ρ 0 , ρ 1 , ρ 2 } and {ρ 0 , μ 1 }, {ρ 0 , μ 2 }, {ρ 0 , μ 3 }.

So the left cosets of K are:

{ρ 0 , μ 1 }, {μ 2 , ρ 2 }, {ρ 1 , μ 3 }

Let’s do the right cosets as well. One right coset is K ρ 0 = K. Now, ρ 1 hasn’t appeared yet, so let’s look for K ρ 1.

Important note: there are four elements you could have picked here, and any one of them is a valid choice.

K ρ 1 = {ρ 0 ρ 1 , μ 1 ρ 1 } = {(1)(2)(3)(123), (1)(23)(123)} = {(123), (13)(2)} = {ρ 1 , μ 2 }.

The two elements not accounted for are ρ 2 and μ 3 , and

K ρ 2 = {ρ 0 ρ 2 , μ 1 ρ 2 } = {(1)(2)(3)(132), (1)(23)(132)} = {(132), (12)(3)} = {ρ 2 , μ 3 }.

So the right cosets of K are

{ρ 0 , μ 1 }, {ρ 1 , μ 2 }, {ρ 2 , μ 3 }

while the left cosets are

{ρ 0 , μ 1 }, {μ 2 , ρ 2 }, {ρ 1 , μ 3 }

These aren’t the same. Except for K , there are lots of partial overlaps.

I hope you paid attention to the logic here. One worksheet exercise on Wednesday will be to do the same thing with the group L = {ρ 0 , μ 2 }.

We can first list the cyclic groups generated by the elements (I’ll ignore the identity ρ 0 .) We have a cyclic subgroup of order 4, since ρk = ρk 1 , and a bunch a cyclic subgroups of order 2.

〈ρ 1 〉 = 〈ρ 3 〉 = {ρ 0 , ρ 1 , ρ 2 , ρ 3 }, 〈ρ 2 〉 = {ρ 0 , ρ 2 }, 〈μ 1 〉 = {ρ 0 , μ 1 }, 〈μ 2 〉 = {ρ 0 , μ 2 }, 〈δ 1 〉 = {ρ 0 , δ 1 }, 〈δ 2 〉 = {ρ 0 , δ 2 }.

If H is a subgroup and ρ 1 ∈ H, then 〈ρ 1 〉 ⊂ H, and that’s already four elements, so if H has any more elements, then |H| ≥ 5. Thus H = D 4. In other words, no other proper subgroup can contain ρ 1 , and the same for ρ 3.

If a subgroup is not 〈a〉, then it has to have more than two elements, and so it has exactly four, the identity ρ 0 and three chosen from {ρ 2 , μ 1 , μ 2 , δ 1 , δ 2 }.

We saw in Monday’s worksheet that {ρ 0 , ρ 2 , μ 1 , μ 2 } is a subgroup. It will turn out that {ρ 0 , ρ 2 , δ 1 , δ 2 } is also a subgroup, but that’s the only other one. Its table looks remarkably similar.

As a reminder,

ρ 0 =

= (1)(2)(3)(4) ρ 2 =

δ 1 =

= (13)(2)(4), δ 2 =

Finally, suppose H is a subgroup of D 4 and H has both a μ and a δ:

μ 1 =

= (12)(34), μ 2 =

δ 1 =

= (13)(2)(4), δ 2 =

It’s not hard to see that

μ 1 δ 1 =

= ρ 3 μ 1 δ 2 =

= ρ 1

μ 2 δ 1 =

= ρ 1 μ 2 δ 2 =

= ρ 3.

Thus, any subgroup that contains μi and δj is forced to contain ρ 1 or ρ 3 , and so is all of D 4.

Thus, the list given is all the subgroups of D 4. You can also find this as Figure 8.13 in the book.