Math 417 – Tenth Day – Class, Lecture notes of Algebra

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Math 417 – Tenth Day – Class

Bruce Reznick University of Illinois at Urbana-Champaign

September 16, 2020

I have gotten several useful emails suggesting that I review some of the ideas of the last lecture. Suppose, for the moment, that G is a finite group. (Most of what I say still applies when it’s infinite, but this will allow us to count.)

Suppose G > H so H is a subgroup of G. We can look at the cosets, which come in two flavors, the left cosets gH and the right cosets Hg. Basically everything we can prove about left cosets can also be proved about right cosets. Some facts:

(i) G can be written as a union of left cosets which are disjoint. The number of cosets, r = [G : H], is equal to |G |/|H|.

G = a 1 H ∪ · · · ∪ ar H; i 6 = j =⇒ ai H ∩ aj H = ∅.

(You can do the same thing with right cosets.) Two left cosets are disjoint, because z ∈ xH ∩ yH =⇒ xH = yH.

(v) Here’s an example. Look at C 8 = {e, a, a^2 , a^3 , a^4 , a^5 , a^6 , a^7 }, with H = {e, a^4 }.

The theorem tells us that C 8 can be written as a union of disjoint cosets. Let’s say my favorite element is a^5. I don’t see it in a coset yet, so look at a^5 H = {a^5 ∗ e, a^5 ∗ a^4 } = {a^5 , a^9 } = {a^5 , a}.

Now I have H ∪ a^5 H = {e, a^4 } ∪ {a^5 , a}. What’s missing? Well, there are four choices: a^2 , a^3 , a^6 , a^7. Pick one, say a^7 , and look at its coset: a^7 H = {a^7 ∗ e, a^7 ∗ a^4 } = {a^7 , a^11 } = {a^7 , a^3 }.

And H ∪ a^5 H ∪ a^7 H = {e, a^4 } ∪ {a^5 , a} ∪ {a^7 , a^3 }. What’s missing? Only a^2 , a^6 , and a^2 H = {a^2 ∗ e, a^2 ∗ a^4 } = {a^2 , a^6 }. We’re done.

C 8 = H ∪ a^5 H ∪ a^7 H ∪ a^2 H = {e, a^4 } ∪ {a^5 , a} ∪ {a^7 , a^3 } ∪ {a^2 , a^6 }.

The order of the cosets and the order of elements in the coset don’t matter: as sets {a^5 , a} = {a, a^5 }, etc.

(vi) If we are looking for subgroups of G , the first step is to look at 〈g 〉 for every g ∈ G. When G is a cyclic group, these are all the subgroups we find. This was also true for S 3 , but not D 4. We can use Lagrange’s Theorem to look at the size of the potential group.

I want to use Lagrange’s Theorem to finish the description of groups of order 6.

What we did the other day was this: Suppose G is a group of order 6 with an element a so that {e, a, a^2 } are distinct and a^3 = e. Suppose b is another element of G and b^2 = e. We saw that either ba = a^2 b (and G is isomorphic to S 3 ) or ba = ab, (and G is isomorphic to C 6 ).

Today I’ll make no hypotheses on G , except that |G | = 6, and show that these are the only possible groups.

Suppose now that G is a group of order 6 and G has no element of order 3 or order 6. I’ll show that this is impossible.

If x ∈ G and x 6 = e, then x must have order 2. Pick such an x. G has 6 elements and {e, x} only gives 2, so there has to be another element in G , call it y , so e, x, y are all different.

What about xy? If xy = e, then xy = x^2 =⇒ y = x. If xy = x = xe, then y = e. If xy = y = ey , then x = e. These impossibilities say that H = {e, x, y , xy } are all distinct. Look at the multiplication table for H. We know that x^2 = y 2 = (xy )^2 = e and, say y (xy ) = (yx)y = (xy )y = xy 2 = x. We can fill out the table completely. H e x y xy e e x y xy x x e xy y y y xy e x xy xy y x e

Another copy of the Klein 4 group!

What could be wrong with that? Well, H is a subset of G and H is a group, so it’s a subgroup of G , and |H| = 4, |G | = 6 and 4 doesn’t divide 6.

What does this mean? It means that our assumption that there were no elements of order three leads to a contradiction, so suppose a ∈ G and a has order 3, {e, a, a^2 } are in G , and let b be another element of G. Then ab, a^2 b have to be in G , because it’s closed under multiplication, and we’ve already shown that {e, a, a^2 , b, ab, a^2 b} are all different.

But now we know something more: we know that b^2 = e or b^3 = e. We did the work a few days ago to handle the case b^2 = e. Now suppose b^3 = e. I won’t need to worry about ba.

THEOREM If G is a group and |G | = 6, then G is either isomorphic to C 6 or isomorphic to S 3.

Since 2,3,5,7 are prime, any group with those orders must be cyclic. We’ve also completely analyzed groups of order 4 and order

6. It turns out that there are five different groups of order 8: we’ve already seen three of them: C 8 , D 4 , and (Z/ 2 Z) ⊕ (Z/ 4 Z). We’ll see the other two eventually.

The number of non-isomorphic groups of a given order can get very large, especially when the order is a power of a prime.

There are 14 non-isomorphic groups of order 16 = 2^4 , 267 non-isomorphic groups of order 64 = 2^6 , 56092 non-isomorphic groups of order 256 = 2^8 and 4948736522 non-isomorphic groups of order 1024 = 2^10.

In fact 99.15% of all the groups of order < 2000 have order 1024.

WORKSHEET PROBLEM 1. (As promised.) Find the left cosets and the right cosets for the subgroup L = {ρ 0 , μ 2 }.

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

Now the right cosets. As always, L = Lρ 0 = {ρ 0 , μ 2 } is one of the cosets. I don’t see ρ 1 so I’ll look at

Lρ 1 = {ρ 0 ρ 1 , μ 2 ρ 1 } = {ρ 1 , μ 3 }.

I don’t see ρ 2 , so I’ll take that:

Lρ 2 = {ρ 0 ρ 2 , μ 2 ρ 2 } = {ρ 2 , μ 1 }.

So the right cosets are

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

Recall, the left cosets were

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

Again, a partial overlap.