## Group Theory Primer, part 5: the last of the isomorphism theorems.

### June 28, 2010

Last time we talked about a whole lot of stuff.  We did homomorphisms, isomorphisms, and talked about the first ismorphism theorem.  What did this one state?  It states that for $G,H$ are groups and $f:G\rightarrow H$ is a homomorphism, then we have that $G/Ker(f) \cong Im(f)$, or, in other words, the quotient of $G$ with the kernel of the map is equal to the image of the map.  This makes sense if you think about it: we’re kind of condensing everything that goes to 0 when we map it away from $G$ and we say that these elements ultimately don’t matter in the image — but, because of the nice properties of homomorphisms, a lot of other elements map onto each other, too.

Today, we’re going to discuss the final two isomorphism theorems (which don’t come up as often, but they’re nice) and conclude with one of the most used theorems in elementary abstract algebra: Cauchy’s Theorem.

Let’s define a product of two subgroups of $G$ before we begin.  If we have $H\subseteq G$ and $K\subseteq G$, then we define $HK = \{hk|h\in H, k\in K\}$.  Yeah, it’s just the product of elements.  For example, in ${\mathbb Z}$ if we take $H = 3{\mathbb Z}$, or the multiples of 3, and $K = 5{\mathbb Z}$, or the multiples of 5, we have that the product of those two will be every element which is of the form $3^{i}5^{j}$ (why?).  Note that in this product, $HK$, we have $H\subseteq HK$ and $K\subseteq HK$.

Now, we know what the intersection of two sets is, right?  $A\cap B$ is the set of elements which is common to both $A$ and $B$; in other words, an element is in $A\cap B$ if and only if it is in both $A$ and $B$.  Okay, so now…

## Second Isomorphism Theorem!

Let’s state the theorem first.

Theorem (Second Isomorphism Theorem): Given $G$ is a group and $H$ is any subgroup of $G$ and $N$ is a normal subgroup of $G$:

1. The product $HN$ is a subgroup of $G$.
2. The intersection $H\cap N$ is normal in H (meaning that $H\cap N$ is a normal subgroup of $H$).
3. We have an isomorphism $(HN) / N \cong H / (H\cap N)$.

Proof. Let’s check (1) first because it’s the easiest.  Is the identity in $HN$?  Yes.  Is it closed under the group operation?  If $a,b \in HN$, this means that $a = h_{1}n_{1}, b = h_{2}n_{2}$ for $h_{i}\in H$ and $n_{i}\in N$.  Then $ab = h_{1}n_{1}h_{2}n_{2}$.  We use a clever trick, by multiplying the inside by an element and its inverse: we have that $ab = h_{1}h_{2}h_{2}^{-1}n_{1}h_{2}n_{2}$, and since $N$ is normal in $G$ this means that $gng^{-1} \in N$ (why?) and so we have $ab = (h_{1}h_{2})(h_{2}^{-1}n_{1}h_{2})n_{2}$ implying $ab = (h_{3})(n_{3}n_{2})$ where we have made $h_{1}h_{2} = h_{3}$ and $h_{2}^{-1}n_{1}h_{2} = n_{3}$.  Obviously $h_{3}\in H$ and $n_{3}n_{2} \in N$, so $ab\in HN$.  Whew.

We must last check to see if inverses are in this group.  If $a\in HN$ then we have $a = hn$ as above.  What is $a^{-1}$ going to be?  $a^{-1} = (hn)^{-1} = n^{-1}h^{-1} = h^{-1}(hn^{-1}h^{-1})$$h^{-1}\in H$ and because of $N$‘s normalcy, the latter factor is in $N$, which means the inverse of $a$ is in $HN$.   Therefore, $HN$ is a subgroup.

Now, there was a lot of manipulating here, but make sure you understood it.  This type of proof will come up time and time again when you have normalcy floating around.

Alright, onto part (2).   We want to show $H\cap N$ is normal in $H$.  This is the same as showing that $h(H\cap N)h^{-1} = (H\cap N)$ for all $h\in H$.  Given an element $r \in H\cap N$ we have that $r\in H$ and $r\in N$.  Then we have that $hrh^{-1}\in H$ obvious, since all the elements are in $H$, but notice also that $hrh^{-1}$ is in $N$ due to its normalcy (since $h\in G$ as well!).  That means $r\in (H\cap N)$.  How nice.  This proves the second part.

The last part, part (3), looks scary, but it isn’t.  Let’s see…we want to make a map between $H/(H\cap N)$ and $(HN)/N$ so that it similar to the first isomorphism theorem with the modding stuff out and all.  It suffices to find a map with kernel $(H\cap N)$ since, if we do, we can apply the first isomorphism theorem and the two quotient groups in this theorem will be isomorphic.

Let’s define a map $\phi: H\rightarrow (HN)/N$ by $\phi(h) = hN$, or, in other words, the coset $N$ multiplied by an element of $H$.  We need to show a few things:

• that this map is a homomorphism.
• that this map is surjective.
• that this map gives us a kernel of $(H\cap N)$.

Once we have these things, we can finish the proof.  This is obviously a homomorphism (check this for yourself!).  Now, we prove that this is surjective.  Suppose that $(hn)N$ is an element of $(HN)/N$ (notice that this is a coset with coefficient in $HN$).  We want to show that some element from $H$ goes into it under $\phi$.  Well, this is kind of convenient: look at $(hn)N$ for a moment.  We have that $n\in N$, so we have that $(hn)N = h(nN) = hN$ (why?).  Since $h\in H$ we have that $\phi(h) = hN = (hn)N$, and so this homomorphism is surjective.

Now, what exactly is the kernel of $\phi$?  That is, what goes to the identity in $(HN)/N$?  Well, what is the identity in $(HN)/N$?  It is simply the coset $N$!  So, we need elements $h\in H$ such that $\phi(h) = hN = N$ which happens if and only if $h\in N$.  Therefore, $h\in N\cap H$, which means that $ker(\phi) = N\cap H$.

Now, we apply the first isomorphism theorem and we obtain the following: $H/ker(\phi) \cong im(\phi)$, but, of course the image of $\phi$ is $(HN)/N$ and the kernel of $\phi$ is $N\cap H$, which implies that $H/(H\cap N) \cong (HN)/N$.  This completes the proof.  $\Diamond$

Note that this proof requires $N$ to be normal in $HN$ (in order to take the quotient), which I did not prove.  This is up to the reader to prove!

## Third Isomorphism Theorem.

Let’s state the result first:

Theorem (Third Isomorphism Theorem): Given $G$ a group and $N, K$ are normal subgroups of $G$ such that $N\subseteq K\subseteq G$.

1. We have $N/H$ is normal in $G/H$, and
2. We have the quotient $(G/H)/(N/H) \cong G/N$.

My algebra professor, Prof. Herrmann, would often call this theorem by an adorable pet name: the freshman theorem.  Why?  Give something that looks like $(G/H)/(N/H)$ to a freshman math student, and he’ll probably guess (since it looks so much like a fraction!) that it’ll reduce to $G/N$.  Cute.

Proof. I’m not gonna get my hands dirty with (1), you’re on your own with that.  It’s really just a matter of showing normalcy, and it is very similar to the manipulating we did in the proof above.

For (2) we have to do a bit of fancy footwork.  Let’s define the natural map $\pi: G\rightarrow G/H$ as $\pi(g) = gH$.  This is called the projection map in some places; it’s just sending $g$ to its coset in $G/H$.  It should seem reasonable, then, that we have a map $f: G/N \rightarrow G/H$ by defining it essentially the same way: $f(gN) = \pi(g) = gH$.  So the only thing that $f$ is doing is telling us to “forget about the $N$” and just send $g$ to where it should go in $G/H$.

We need to prove that $f$ is surjective.  If we have some $gH\in G/H$, then we have $g\in G$ and so we have $f(gN) = \pi(g) = gH$.

Now we need to ask ourselves what the kernel of $f$ is.  Well, if we have $f(x) = H$, then this implies that the elements $x$ that satisfy this are exactly those $x$ such that $x\in G/N$ and such that $xH = H$.  In other words, we have that $x\in H/N$ implies that $f(x) = H$.  Check this out, it makes sense.  You may need to think about these last few sentences for a bit: they’re not ridiculously obvious.

Now that we have the kernel for $f$ we note, by the first isomorphism theorem, that $(G/N)/ker(f) \cong im(f)$.  Since $f$ is surjective, $im(f) = G/H$, and since the kernel is $H/N$, we have that $(G/N)/(H/N) \cong G/H$.  This concludes the proof.  $\Box$

## A Cauchy Break!

These last two theorems were much more tedious than the first isomorphism theorem, and, in fact, there is a fourth isomorphism theorem!  I may write about that, but I probably won’t prove it.  It’s kind of dull.

The kinds of proofs we did above are standard kinds of proofs: we run into this kind of thing all the time.  With normalcy, we ordinarily use the trick where $1 = h\cdot h^{-1}$ and insert terms cleverly into other terms, or see if we can get things into the form $gNg^{-1}$ somehow.  With cosets, we often use projection mappings.  With isomorphisms, we ordinarily try to find a map with a nice kernel and use the first isomorphism theorem.

I’m going to conclude this post with the statement of cauchy’s theorem and one application of it.   It’s pretty cute, no lie.  Let’s define two terms before we start: the order of a group is simply how many elements are in the group.  The notation is just $|G|$.  For example, if $G$ has 9 elements, then $|G| = 9$.  The order of an element in a group is a little different; for an element $g\in G$, it is the smallest number $n$ such that $g^{n} = 1$, where 1 is the identity of $G$.  Confusingly enough, it is denoted the same way: as $|g|$.  For example, in the $D_{8}$ (the one where we rotated the square around) we have that $|r| = 4$ and $|s| = 2$.  Note that $|D_{8}| = 8$, which is why we call this group the dihedral group of order 8.  Hope everything’s comin’ together for you.

Theorem (Cauchy’s Theorem): Let $G$ be a finite group and $p$ is a prime.  If $p$ divides the order of the group, then there is an element of order $p$ in the group.

So what does this state?  Suppose that $|G| = 21$.  Well, since 3 and 7 go into 21 and 3 and 7 are primes, we must have an element of order 3 and an element of order 7 in $G$.

Think about $D_{8}$ for a moment.  As we noted above, $D_{8}$ has order 8, and the only prime that goes into 8 is 2.  Do we have an element of order 2?  Yes, we have $s$.  Note that this is not necessarily unique, since the element $r^{2}$ also has order 2 (since $(r^{2})^{2} = r^4 = 1$).

We will not prove cauchy’s theorem, but we will use it to prove another cute theorem.  Let’s define one more term before we pack up our things:

A group is cyclic if it is just the multiples of some element (usually called the generator).  For example, if we take the number 2 and we keep taking multiples of it, then we will have a cyclic group.  If we work in mod 12 arithmetic (clock arithmetic) we can take the element 3 and take all the multiples of it: we get 3, 6, 9, 12 = 0, 3, 6, 9, 12 = 0, … and so on.  This means that $\{0, 3, 6, 9\}$ is a cyclic subgroup of ${\mathbb Z}_{12}$, which is the proper name of the integers mod 12.

Another cyclic group is something like:  $\{a^{n} | n\in {\mathbb N}; a^{3} = 1\}$ under multiplication.  What is this group?  Well, we have the elements $\{a, a^2, a^3, a^4, a^5, \dots\}$, but we know that $a^3 = 1$ so we can reduce these elements to $\{a, a^2, 1, a, a^2, 1, \dots\} = \{1, a, a^2\}$.  This is also a cyclic group, and we say that $a$ generates it, since we’re taking multiples of $a$.

Now, the theorem:

Theorem:  Given $G$ a finite group such that $|G| = n > 1$ and $p$ is a prime which divides $n$, then $G$ has a cyclic subgroup which has order $p$.

Proof. Think about this one yourself for a bit.  Think of Cauchy.  When you’re tired of thinking, or just wanna cheat, keep reading.

By Cauchy, since $p$ divides $n$, there exists an element $g\in G$ such that $|g| = p$.  We’re going to let $g$ generate a subgroup.  Let $H = \{g^{n}| n\in {\mathbb N}\}$.  We have $g^{p} = 1$ by assumption, and this is the smallest such number such that $g^{n} = 1$, which means that $H = \{1, g, g^2, \dots, g^{p-1}\}$.  Therefore, $H$ has order $p$ and is the subgroup we required to prove this theorem!  $\Box$

## The End of Groups as We Know it?

Certainly not!  But this is the last of my group theory primers.  For the types of things I’d like to cover in this blog there is not all that much more to know.  Furthermore, if there are new things to know, I will want to introduce them a bit before we actually use them (like free groups before we go over certain fundamental groups).  Without being able to apply these things to anything, you will quickly forget them; this is why I didn’t want to go step-by-step over every little thing at the beginning of a normal abstract algebra class.  Either way, I highly recommend seeking the comforting questions tucked deeply within the pages of your favorite abstract algebra book: I use Dummit and Foote’s book, but I.N. Herstein is just as good.  A much easier book is Frayleigh’s algebra textbook.  Whatever book you have on abstract algbera is probably fine, and if you don’t have a book on abstract algbera, you’re going to have to be a clever little fox on the internet and find one.

So, what was the point of these primers?  If you remember (or if you’re a new reader), the point of these was to use group theory to help us understand Linear Algebra.  I wanted to be able to push things back and forth between matrices and groups and go over representation theory.  This is a fun goal, but I feel I should modify it a bit.  I feel that I should introduce my readers slowly to the theorems of representation theory, but focus on a primer to point-set topology.  I have already written a primer on set theory, which I will eventually post with little modification.

The proofs of point-set topology will rely heavily on set theoretical notions, especially in the very beginning when we need to build up our machinery.  Any readers who are not comfortable with set theory at the moment will be happy to know that all of the set theoretic concepts I use will be defined at some point in this blog.  The interested and set-savvy reader should really start with Munkres (after the set theory chapter) for a great (if not a bit dry) introduction to point-set. I will be describing the same things, but I will emphasize slightly different aspects and I will, of course, be covering much less ground.  If anyone has any suggestions or things they’d like to see covered, I’m open to it!