I’m Sorry

I’m sorry. I know I said I wouldn’t make any posts about math but this is the only one, I promise. Anyway, my motivation for writing this was that Group Theory is considered one of the hardest undergraduate math classes. But the ideas in introductory group theory are actually very simple. And moreover, in a standard undergraduate course, there aren’t even that many ideas. What makes group theory hard is that there is so much new terminology being thrown out that it can be hard to decipher what is even being said.

To the newcomer, learning group theory is kind of like trying to read Jack and the Beanstalk in French with only a French to English dictionary. You would understand each word individually but by the last word in a sentence you likely would have forgotten what the sentence was about in the first place. So even though the story of Jack and the Beanstalk is not that complex, you might finish the story with not idea about what happened at all. Learning group theory, there are so many new terms that you have to constantly be flipping to different pages to your notes to reference different definitions and in the process you lose sight of what is really happening. (Group Theory is definitly not the only area of mathatics like this. I think its just one of the worse offenders)

Really, the only answer to this problem seems to be just doing a lot of excersises until you eventually internalize all the definitions but I would like to try and cheat. It probobly won’t work and trying to not include terminology will just make everything harder to understand but I have nothing better to do with my time.

So these are my sparse notes containing as little terminology as possible and only focusing on the fundamental ideas. Really, there are only a couple ideas needed to understand group theory at this level but these ideas are often lost in traditional texts, shrouded by mountains of terminology and theorems that are actually obvious if you understand the fundamental ideas. So if you can get through these notes and really understand them, then I think you should be able to learn all the rest of undergraduate group theory very quickly.

These notes are written at a basic level so anyone that has taken precalc should be able to understand everything. This does mean I simplified some stuff a little which to a mathematician would probably be heresy but I don’t think it will affect your understanding at all.

I alluded to this above but I wan’t to make explicit these are very short notes and nowhere near all of a standard course in Group Theory. Just what I think is the most important and a very solid foudation for understanding everything else.

What is a Group?

A group is set together with a binary operation *. A binary operation just means that you can combine any two elements to get another element. Addition and Multiplication of numbers are both binary operations. As we’ll see lots of sets of numbers are also groups. Anyway, in order to be a group the binary operation must satisfy the following requirements.

  1. Associativity: (a*b)*c = a*(b*c)
  2. Identity: There exists a unique element of the group called the identity, we will always refer to it as 1, so that for any element of the group a, we have 1*a = a*1 = a
  3. Inverse: For any element of the group a, there exists a unique element of the group a^{-1} that satisfies a*a^{-1} = a^{-1}*a = 1.

This is a lot to digest so let’s spend some time understanding what this actually means. First, from now on we will refer to the binary operation as “Multiplication” This doesn’t mean multiplication in the traditional sense you are used to. We just use it because we want a shorthand for “using the binary operation to compose two elements.” To avoid confusion, I will write traditional multiplication if I mean the traditional multiplication of numbers you are used to.

Associativity just means that the order of how you multiply doesn’t matter. This means we eschew parenthesis so we will write (a*b)*c as a*b*c and we will actually just drop the * because it is annoying to write. So it will just be abc.

In addition, 0+ \text{something} = \text{something}. In traditional multiplication 1\times \text{something} = \text{something}. The identity is a way to generalize this idea to groups.

In addition, \text{something} + -\text{something} = 0. In traditional multiplication, \text{something} \times \frac{1}{\text{something}} = 1. (As long as “something” is not zero of course). Again, inverses are a way to generalize this idea to groups.

Notice that communitivity is not one of the requirements for something to be a group. This means that in some groups it is not always true that ab = ba.

Lets look at some examples.

Example 1. Real numbers where the “multiplication” is addition. Here, 0 is the identity and negatives serve as inverses. (This might be confusing because we usually denote the identity as 1 but in this case it is 0 and 1 is just some random element with not special importance.

Example 2. Real numbers where the “multiplication” is traditional multiplication. Actually, this not technically a group. (why?) Because, while 1 is an identity and \frac{1}{a} is the inverse for most a, notice that there is no inverse for zero. There is no number that we can multiply zero by to get 1. Okay, so the real numbers using traditional multiplication is almost a group but not quite. But if we just removed zero our problems are solved. So the real numbers without zero is a group where the “multiplication” is traditional multiplication.

Example 3. The integers where the “multiplication” is addition.

Okay, these past examples are all things you are probably familiar with and they turn out to be groups. Next were going to get into groups that are really close to numbers but you might not have seen it before, after that we will look at examples of groups that are nothing like numbers. Additionally, from now on, all groups we will examine will be finite. And assume all groups I am talking about are finite. (Being finite just means having finitely many elements).

Example 4. Mod 5 where the “multiplication” is addition. Mod 5 means the elements of the group are the non-negative integers less than 5, and if you would add two numbers and get something greater than 5, you take the remainder modulo 5 instead. So for example,

3+4 =2


3+2 =0


2+2 =4

The identity is 0 and the inverse of something is 5-something (unless something is 0, in which case the inverse of 0 is 0). So the inverse of 4 is 1 and the inverse of 3 is 2. Obviously “5” can be replaced by any positive integer and you will be left with a group.

Example 5. Mod 5 where the “multiplication” is traditional multiplication and the elements are 1,2,3,4 (notice 0 is not an element of this group). This is the same as mod 5 with addition except you multiply instead of add. So for example.

3\times 2 = 1

4\times 3 = 2

4 \times 2 = 3


3\times 1 = 3

The identity is 1. The existance of invereses might not be obvious but notice that is sufficies to see that for any a,b,c

ab = ac

implies b=c (why?) But notice that ab = ac is equivilant to

a(b-c) =0

which can only happen if b = c. Notice that mod n does not always form a group under multiplication though. For example in mod 4, 2 does not have a multaplicative inverse. See if you can find necissary and sufficient conditions for mod n to be a group with traditional multiplication.

Now we’re going to see some groups that don’t look anything like numbers at all.

Example 6. Permutation on 4 elements. The elements of the group are the permutations of 1,2,3,4 which recall are just, 1234, 1243, 1423, 4123, 1324, 1342, 1432, 4132 ect. (You get the idea). We can interpret a permutation as one-to-one correspondence between the set \{1,2,3,4\} and itself by interpreting the permutation where a i is in the jth place as f(j) = i. So for example, 1432 would be the function where

f(1) =1

f(2) = 4

f(3) = 3


f(4) = 2

And 4321 would be the function where

f(1) = 4

f(2) = 3

f(3) = 2


f(4) = 1

hopefully you get the idea. Multiplication is just function composition so

(4321)(1432) = 4123

This is the first example where the multiplication isn’t commutative; notice that

(1432)(4321) = 2341

The identity is the identity function 1234 and the inverse is just the inverse function. For example the inverse of 1432 is 1432.

Playing With Groups

Lets just tinker around a little to get a feeling for how groups behave.

Suppose that in some group

ab = ac.

We can multiply both sides of this equation on the right by a^{-1}. So we’re left with

b = c.

This is an important example of a property of groups and that is that by using inverses, we can essentially cancel two terms that are the same on both sides of an equation.

Subgroups: Start with a random group G and some a that is an element of G. What if we multiply a by itself? Then we get another element of G: aa. We’ll call this a^2. Okay lets multiply that by a. Now we’re left with aaa which we’ll call a^3. And we can keep doing this generating a sequence a, a^2 \cdots. Now, remember earlier we said we will assume that groups are finite. This means that at some point our sequence must start repeating. So we have a^j = a^k for k > j. Using the cancellation we saw above, we are left with

a^{k-j} =1

Assuming that there is no positive integer less than $k-j$ satisfying a^n =1, we see that the elements 1, a, a^2, \cdots a^{k-j-1} actually forms a group on its own. And this looks awfully similar to mod k-j. This is what we call a subgroup. A subgroup is a subset of a group that itself forms a group. Not all subgroups arise by multiplying an element by itself though. For example, if you start with any subset of elements  g_1, g_2, \cdots g_n of a group, then the set of all elements you can get by only multiplying together elements from g_1, g_2, \cdots g_n will form a subgroup. (If it is not clear I meant with replacement. So {g_1}^2 or even g_2g_1{g_2}^4 would be in the subgroup)

Using the notation above, what is (a^3)^4. Well, this is just a^3 a^3 a^3 a^3 and that is just

aaaaaaaaaaaa = a^{12} = a^{3\times 4}

so we see that the exponent-like notation for groups works the same way we are used to with numbers. Similarly, (a^n)^{-1}a^n = 1 so multiplying both sides on the right by (a^{-1})^n yields

(a^{-1})^n = (a^n)^{-1}

which we will write as a^{-n} like we are used to with numbers.

If we’re given two groups, can we smash them together to make one big group? Of course we can. For two groups G and H, we define the catesisan product of G and H to be the group G \times H. In other words for any g in G and h in H, we can put them together to make an element g \times h of G \times H. See if you can figure out what the multiplcation is before I tell you. Okay, here it is, for g_1, g_2 in G and h_1, h_2 in H,

(g_1 \times h_1)(g_2 \times h_2) = (g_1g_2 \times h_1h_2)

Verify that this is in fact a group.

Understanding the Homomorphism

So far we have looked at individual groups. Now, we turn our attention to examining how we can find similarities between different groups. To do this we use functions.

We could study any old function between groups but if we don’t put any contraint on them then we’re not using the fact that these are groups. They might as well just be sets.

Instead we focus on Homomorphisms. A Homomorphism is a function between groups that preserves the group structure. What does this mean? Well since we wanted to study similarites between different groups, we want to look at functions where you can multiply stuff and then apply the function, or apply the function and then multiply stuff and end up with the same result.

Fomrally, a Homomorphism is function f between two groups satisfying.

f(a)f(b) = f(ab)

Just this constraint makes sure that the function preserves a lot of stucture. For example notice that

f(1)f(a) = f(1a) = 1f(a)

And we can use the cancellation property we saw above to deduce that

f(1) =1


f(a)f(a)^{-1} = 1 = f(aa^{-1}) = f(a)f(a^{-1})


f(a)^{-1} = f(a^{-1})

Next, lets start with an observation. If f(a) = 1 and f(b) = 1, then f(ab) =1. Aditionally, if f(a) = 1 then f(a^{-1}) = 1. Therefore the set of all elements of a group that get mapped to 1 by a homomorphism form a sub-group.

Now, suppose that this subgroup is a_1, a_2, \cdots, a_n. Notice that for any element b in the group,

f(a_1b) = f(a_1)f(b) = f(b)

And the same is true for any of the a_i. The converse is also true. Suppose that f(b) = f(c) = d. Then we have

f(b) = f(c)= f(bb^{-1}c) = f(b)f(b^{-1}c)

Therefore, f(b^{-1}c) =1. In particular this mean that exactly n elements of our group get mapped to d. Since d was arbitrary, this means that for any element that can get mapped to by our homomorphism, there are exactly n elements of the group that map to it.

We can also partition the elements of a group into Equivalence Classes by what they map to for some homomorphism. What we just saw above is that each of these equivalence classes are the same size and that two elements a \text{ and } b are in the same equivalence class if and only if

f(ab^{-1}) =1

Or to put it into perhaps simpler terms, there exists a c such that f(c) =1 and ac =b. We call these equivilance classes the Quotient of the group by the subgroup. This is written G/H for a group G and subgroup H.

So really what a Homomorphism does, is take a subgroup and wraps it up into one element. Then other elements of the group also get wrapped into one thing if they “differ” by some element of the subgroup.

For example, take the homomorphism that maps mod 10 into mod 5 by f(x) = x for x < 5, and f(x) x-5 for x 5 \leq x < 10. Or put more simply, just takes mod 5. The subgroup that maps to 0 is $\latex \{0, 5\}$, and the equivilance classes are \{0, 5\}, \{1, 6\}, \{2,7\}, \{3, 8\}, \{4, 9\}. See how the homomorphism is taking mod 10 and squishing it down to mod 5.

What happens if no squishing is taking place? What if, for some homomorphism, f the only element that maps to 1 is 1. And moreover ever element of the group f maps to can get mapped to using f. Then nothing is really happening at all. The group getting mapped to and the group getting mapped from are basically the same. f is just renaming elements. We’re just taking an element a and disguising it as f(a) but it still behaves exactly like a. If you put glasses on a cat, its still a cat. In this case we say the homomorphism is an Isomorphism. And if there exists an isomorphism between two groups, we say the groups are Isomorphic. 

Always rememebr, Isomorphic basically just means the same. No squishing.

Wouldn’t it be nice if the quotient of a group by a subgroup was a group? Unfortunately. If the subgroup is H, then we would like there to exist an h_3 in H so that

ah_1bh_2 = abh_3

for any a and b in G and h_1, h_2 in latex H$. Certaintly it would be true if the group were communative but remember that groups don’t necissarily have to be. We don’t even need communativity, we just need to slide that b to the left of the h_1 and we don’t mind if in the process we turn the h_1 into a different element of H. What we need is there to exist and h_4 in H so that

h_1b = bh_4

or more concisly, for any h \in H and b in G,

b^{-1}hb \in H

This allows us to slide that b right past the h. In this case we say the subgroup H is Normal in G. Just like in real life, being normal is a desirable property.


How can we use group theory in our everyday lives? We can’t, but it can be useful in other areas of mathematics. One of these ways is with actions.

We can think of an action as a way to multiply elements of a group by elements of a structureless set which will in turn give structure to the set. It satisfies some properties to make sure being a group matters:

An Action of a group G on a set S is a function * from G \times S into S (which will be written g*s instead of *(g,s) because that’s annoying to write) which satisfies for any g and h in G and a in S.

  1. g(h*a) = (gh)*a
  2. 1*a = a

Actions are used to solve the Rubik’s Cube. See math is valuable in real life.

I won’t talks anymore about actions beyond stating their definition because most of their introductory theory is very similar to what we have seen already with Homomorphisms. The excersises will provide a way to get an understanding of actions.


These excersises might be too hard but I couldn’t find easier ones so deal with it.

1.  Show that the intersection of two subgroups is a subgroup.

2. Find necissary and sufficient conditions for mod n with traditional multiplication to be a group.

3. (Legrend’s Theorem) For a group G and subgroup of G, H. Show that the number of equivilance classes in the quotient of G by H divides the number of elements in G.

4. The order of an element a of a group is the smallest positive integer n such that a^n =1. Show that the order of each element divides the number of elements in the group.

5. If a group has an even number of elements show that the number of elements of order 2 is even.

6. Show that if f is a homomorphism from G_1 to G_2 and g is a homomorphism from G_2 to G_3 then the composition g of is a homomorphism.

7.  A group G is abelian if its elements commute i.e. ab = ba for all a,b in G. If G is an abelian group and its elements are a_1,a_2, \cdots a_n, show that

(a_1a_2\cdots a_n)^2 =1


8. Prove the Second Isomorphism Theorem:

Let G be a group. Let S be a subgroup of G, and let N be a normal subgroup of G. Then the following hold:

  1. The product SN is a subgroup of G,
  2. The intersection S \cap N is a normal subgroup of S, and
  3. The quotient groups (SN) /N and S/(S\cap N) are isomorphic.


9. Suppose we have a group G, a set S, and an action * of the group on the set. Show for any x \in S that

|G| = |\{g \in G: g*x =x\}| |\{g*x: g\in G\}|

(Note that |A| denotes the number of elements in a set A)

10. Suppose we have a group G, a set S, and an action * of the group on the set. An orbit of S under the action * is a set of the form |\{g \in G: g*x =x\}|where x \in S. If O is the number of distinct orbits of S, Show that

|O| = \frac{1}{|G|} \sum_{g \in G} |\{x \in S: g*x = x\}|

11. (Hard) Prove Cauchy’s Theorem: If p is a prime number and p divides the number of elements in a group G, then there exists a subgroup of G with exactly p elements.

12. (Hard) Prove Sylows 1st theorem: If p is a prime number and $p^k$ divides the number of elements of group G, then the number of elements.

13. (Hard) Show that all Abelian groups are isomorphic to a unique product of cyclic groups.

14. (Really Really Really Hard) Show that if G, H_1,H_2 are finite groups and G \times H_1 is isomorphic to G \times H_2 then H_1 and H_2 are isomorphic.

(Funny/Infuriating Story: When I took this course last year, the teacher was trying to prove a theorem. I sketched a proof in my head using this claim (I thought is should be simple to prove and I had a sketch of the claim in my mind) and went back to doing something else. Like 30 minuets later I looked up and the professor was still doing the same proof. He was struggling and looked confused I raised my hand and offered my outline. The professor told me it wouldn’t work because this claim wasn’t true. I was puzzled because it seemed so intuitivly obvious. I tried to actually prove it but realized that the obvious technique did not work. After class I asked the professor if he could explain to me why it wasn’t true or offer a counterexample. He told me it wasn’t true because the obvious method to try to prove it didn’t work! He even went through the work of doing the obvious method which I told him I had already tried and when it didn’t work he exclained “See this doesn’t work. The claim isn’t true.” This was a a professor of mathematics. I went home and looked up the claim online and sure enough it was true.)


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