One of the ways of producing (potentially) new groups from old is called the direct product.Let me show you the direct product of two groups by example, first.
Let me invoke the abstract algebra package and set it to “groups”.
C2 x C2
I want two copies of C2, the cyclic group of order 2…
… and I am going to ask for their “direct product”.
What did we get?
First of all, we see that the elements of the direct product are just ordered pairs – the first component from H, the second component from K.
Second, looking at the Cayley table, we see that the product is simply the componentwise products: {x,y} {X,Y} = {xX, yY}.
Third, there’s a standard name for the resulting group: the Klein Fourgroup. The Abstract Algebra package knows it as Klein4:
Things really are the same. Each group is abelian and has three elements of order 2:
… and I even ordered the elements so that (a,1) is paired with a, etc. Note that I am going to write the ordered pairs using parentheses, although Mathematica uses curly braces; I will want to distinguish, in particular, the ordered pair (1, b) from the set {1, b}.
We can construct the direct product using ordered pairs… but once we’ve got it, we can drop the commas and use multiplication. We match (1,1) with e… (a,1) and (1,b) with a and b respectively… and (a,b) with c. the Klein 4group is {1, a, b, c} with a^2 = b^2 = c^2. Furthermore, c = ab = ba. This matchup is an isomorphism, between the Klein 4group and C2xC2.
Note that H and K are not, properly speaking, subgroups of HxK – but they are isomorphic to subgroups (H,1) and (1,K), by which I mean and . In words, (H,1) is the set – subgroup! – whose first component is from H and whose second component is 1.
On the other hand, H and K are not only subgroups of the Klein 4group, but normal subgroups (because the Klein 4group is abelian)… and their intersection is the identity: … and the product of the orders of H and K is equal to the order of the Klein 4group, G:
H K = 2*2 = 4 = G.
It turns out that whenever we have those properties, we can conclude that HK, the set of products of the form hk, is isomorphic to the direct product HxK.
That is, let G be the Klein 4group {1, a, b, c}… let H = {1,a} and K = {1,b} (those are sets, now, not ordered pairs). We have H and K normal in G, and , so
HK ~ HxK
but also HK = H K = 4 = G, so
HK = G
so
G ~ H x K
(where I use ~ to mean “isomorphic”).
These two, ordered pairs and ordered products, are called the external direct product and internal direct product, respectively.
One last thing before we move on to another example.
is C2 x C2 ~ C4 ?
No: C4 has an element of order 4, while C2 x C2 does not. The orders of elements is the first thing to check when wondering about isomorphism between two groups. That they all match is not proof of isomorphism – I must know a counterexample… when I think of it – but if any do not match, then we cannot have isomorphism.
So, in this case, the direct product C2 x C2 gave us a new group of order 4.
C2 x C3
Now let me form the (external) direct product C2xC3 of C2 and C3…
That’s symmetric about the main diagonal, hence the group is abelian. It’s of order 6… what about the abelian group of order 6 we already know about, namely C6?
That looks prettier, and different. But let’s look more closely.
There’s nothing in the orders of the elements to rule out the groups being isomorphic… let’s see if we can construct an isomorphism.
There’s only one element of order 2 in each group, so one should be mapped to the other: (a,1) should be mapped to g^3.
We have two elts of order 6, so (a,b) and (a, b^2) should be mapped to g and g^5. But which way? Does it matter? What are the powers of (a,b) and of (a, b^2)?
Wait a minute. Each of those lines shows that either (a,b) or (a, b^2) generates the entire group of 6 elements. We just showed that C2xC3 is a cyclic group of order 6 – but there’s only one, up to isomorphism. We don’t even need to actually construct an isomorphism.
But let’s look at the two corresponding lines for C6 (for some reason I had to ask for Element[[2]] rather than “g”):
I think I want (a,b) mapped to g… and I’m going to check that the powers of (a,b) match the powers of g:
Take another look at J:
The order is 1, 5, 3, 4, 2, 6… so let me change the order in the Cayley table for J – that was the point of these most recent calculations – and show them sidebyside:
OK. The colors match, and tell me that I have an explicit isomorphism. We have seen again that the external direct product C2xC3 is isomorphic to C6.
What about the internal direct product? I believe I can find normal subgroups of C6, of orders 2 and 3, intersecting in {1}.. Well, there’s only one element of order 2, namely g^3, so I let H = {1, g^3}; then K is everything else (plus the identity): K = {1, g^2, g^4}.
H and K are subgroups of an abelian group, so they are normal. All I need to do now is write every element of C6 as a product hk:
1 = 1 1
g = g^3 g^4
g^2 = 1 g^2
g^3 = g^3 1
g^4 = 1 g^4
g^5 = g^3 g^2
or do it the other way:
1 1 = 1
1 g^2 = g^2
1 g^4 = g^4
g^3 1 = g^3
g^3 g^2 = g^5
g^3 g^4 = g
So I have exhibited the internal direct product.
Finally, we see that C2xC3 is isomorphic to C6, although C2xC2 was not isomorphic to C4. What’s the difference? The orders 2 and 3 are relatively prime – theur greatest common divisor is 1; 2 and 2 are not relatively prime: their GCD is 2. In the external direct product, the 2cycles and the 3cycles only line up after 6 ordered pairs; in the case of C2xC2, the 2cycles line up both times.
In this case, then, we did not get a new group.
This is a general theorem: Cm x Cn is isomorohic to Cmn if and only if m and n are relatively prime.
D3 x C2
There’s no reason to restrict ourselves to cyclic groups. D3 is the smallest dihedral group, and C2 is the smallest cyclic group, so let’s try D3xC2. (Hmm. I know it doesn’t matter in what order I do the factors… but prove it? Well, the obvious map that interchanges components is an isomorphism.)
We got a group of order 12.
Well, the dihedral group of order 12 is D6:
Let’s look at the orders of the elements…
Each has two elements of order 6… two elements of order 3… so we do not rule out the possibility that D6 is isomorphic to D3 x C2.
Here are the Cayley tables, but the colors don’t match. We would have to work at an isomorphism.
They are, in fact, isomorphic, and almost any text should have the proof (it’s more complicated than I like for now). There’s just one caveat: if we write Dn x C2, the theorem requires n to be odd. So any text should prove the isomorphism in general, for Dn x C2, with n odd, rather than just for D3 x C2 (i.e. the case n=3).
D4xC2
At the very least, I should close by looking at D4xC2; I have said that it should not be isomorphic to D8.
Here’s D4… and the direct product with C2:
Compare the orders of the elements of G with the orders of the elements of D8:
The (nonidentity) elements of the direct product have orders 2 and 4; D8 has three elements of order 8. These two groups cannot be isomorphic. So we know that Dn x C2 is not always isomorphic to D2n; as I said, n must be odd.
building groups using direct products
Well, I’m done with examples… but we have just constructed new groups. C2xC2 was a 4element abelian group, a group of order 4, different from C4. D4xC2 was a 16element group different from D8, also of order 16.
When we look at direct products of cyclic groups, remember that Cm x Cn is not isomorphic to Cmn whenever m and n are relatively prime. (I’m going to use symbols for equaity, = and ≠, because they’re convenient, but they stand for “isomorphic” and “not isomorphic”, respectively.
We always have
Cn for any integer n.
Then
C2 x C2 ≠ C4
C2 x C3 = C6
C2 x C4 ≠ C8
C2 x C5 = C10
C2 x C2 x C2 ≠ C8
C2 x C6 = C2xC2xC3 ≠ C12
C2 x C7 = C14
C2 x C8 ≠ C16 ≠ C2 x C2 x C2 x C2 ≠ C2 x C2 x C4
…
C3 x C3 ≠ C9
C3 x C4 = C12
C3 x C5 = C15
…
C4 x C4 ≠ C16
…
C2xD3 = D6
C2xD4 ≠ D8
C2xD5 = D10
…
C2xQ
That is, we have
one (abelian) group of any prime order p:
 Cp
two of order 4:
 C4
 C2xC2
two of order 6:
 D3 (= S3)
 C6 = C2xC3
five of order 8:
 C2xC2xC2,
 C4xC2,
 C8,
 D4,

Q.
two of order 9:
 C3xC3,
 C9
two of order 10:
 C2xC5 = C10,
 D5.
five of order 12 (one of which we have not seen):
 C4xC3 = C12,
 C2xC2xC3 = C2xC6,
 D6,
 A4 (the alternating subgroup of S4)
 and there exists a fifth a group which is a semidirect product of C3 and C4
two of order 14:
 C2xC7 = C14,
 D7.
five abelian groups of order 16:
 C16
 C2 x C8
 C2 x C2 x C2 x C2
 C4 x C4
 C2 x C2 x C4
and we know three nonabelian groups of order 16:
 C2 x Q
 C2 x D4
 D8
There are, according to a table I have, 14 groups of order 16, 6 more than I have shown. In fact, the number of groups of order 2^n grows exponentially. I’m sure I’ll show you some of the groups of order 16, but I don’t know if I’ll show you all of them. (I know exactly one relatively elementary reference. How elementary? That’s the question.)
Getting all the abelian groups of order n turns out to be easy: they are precisely all the possible direct products of cyclic groups. This might be called the Fundamental Theorem, or the Decomposition Theorem, of Finite Abelian Groups: every finite abelian group is the direct product of cyclic groups (possibly only one) . And finding them is fun. I’ll show you.
Getting all the other groups of order n is challenging. It requires what are called the Sylow Theorems, and usually a fair bit of calculation. I doubt that I will do them on this pass through group theory.
February 28, 2012 at 2:03 am
Hello there, I’m new reader of your blog. It is very interesting.
Could you please say what programming language do you use?
Looks like python to me…
Cheers,
Gregory
February 28, 2012 at 2:15 pm
Hi Greg,
I’m using Mathematica® from Wolfram and a free addon package for Abstract Algebra. I believe there is a fullfeatured homeuse version of Mathematica at a reasonable price.
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