# A note on isomorphisms

We’ve now defined an isomorphism. Before we start, recall what this was. Two isomorphic groups must have the same number of elements. From the second exercise in the previous post, we have that isomorphic groups must have the same number of elements of order m for all positive whole numbers m, and the same number of elements of infinite order. However this is a necessary but not sufficient condition. I want to show you two finite groups

• which have the same order (the same number of elements) and
• whose non-trivial elements all have order 3

but are not isomorphic. For one of the examples, we’re going to need direct products, and for the other example we’ll need matrices and fields.

A brief aside: fields. I think it’s often useful to have an example in mind when you’re thinking of a definition or property, as you can then check that your definition ‘fits’ with this idea and sometimes the example can help you to show you what’s special about the property. My thought on a field is the real numbers, denoted $\mathbb{R}$, and the rational numbers (i.e. the fractions), denoted $\mathbb{Q}$. With addition these are both groups.

So what makes a field? Well, a set F becomes a field if

• we have two ways of combining elements, often denoted + and ×
• both (F, +) and (F\{0}, ×) are commutative groups
• these + and × ‘play nice’ in the way they interact, specifically distributivity.

I’ll show you explicitly the axioms using $(\mathbb{Q}, +, \times)$.

• let + and × be the usual operations we know and love.
• then $(\mathbb{Q}, +)$ and $(\mathbb{Q}\setminus {0}, \times)$ are commutative groups (check this).
• we require, for all a, b, and c in $\mathbb{Q}$ that a(b+c)=ab+ac, which we know holds for these everyday operations + and ×.

You may be wondering where the ‘exclude 0’ comes from. I’ve been a bit lax here, as what I really mean by 0 is the identity element from the group (F, +). This is because such an element will always stop (F, ×) from being a group. Why? Well let’s consider the distributivity condition and, for convenience, set 0 to be the identity of (F, +). Now

a(0+0)=a0 + a0 [by distributivity]

a(0)=a0 + a0 [using that 0 is the identity in (F, +)]

a0 + (-a0) = a0 + a0 + (-a0) [where (-a0) is the inverse of a0]

0 = a0

which means that 0 has no inverse in this group. For any a in F, we have that a0 = 0. If this proof looks a bit abstract, look through it but imagine we are in $\mathbb{Q}$. The justification for each step definitely works in this set, and then note that all of the properties we are using are within the axioms for a field, and so they work there too. Note that, since a group must have at least one element, we have that a field must have 2 elements (as F\{0} cannot be empty).

Some more examples of fields. So are $\mathbb{Q}$ and $\mathbb{R}$ the only examples of fields? These were infinite commutative groups. Recall our visualisation of the cyclic group using modular arithmetic (or see here). This gives us a finite group that uses the integers, and so we can be ‘hands on’. Before continuing: when does including multiplication with modular arithmetic give us a group? Use this to discover some finite fields. I’ll go through one below if this is unclear.

A bigger problem would be work out for exactly which n the set {0, 1, …, n-1} with modular addition and modular multiplication forms a field. It is then called the field of n elements, denoted $\mathbb{F}_n$. As mentioned above, a field must have at least 2 elements. So I’ll show that $\mathbb{F}_2$ and $\mathbb{F}_3$ are fields (have a go first if you’d like though).

Ok. So for the field of 2 elements we’re going to use {0, 1}. We then have that ({0, 1}, +) is isomorphic to C2. This part will always be fine (i.e. a cyclic group of order n). So the question is of the other 2 axioms. Let’s check about ({1}, ×) with × representing modular multiplication. Well…not much to check really, this is the group with one element i.e. 1 × 1 = 1. Our final axiom to check is whether a(b+c)=ab+ac. We don’t actually ever need to check 0: our manipulation above will always deal with that. Then 1(b+c)=(b+c)=1b+1c. Done! Now for the field of 3 elements. Here we need to check axioms 2 and 3: is ({1, 2}, ×) a commutative group and do the operations ‘play nice’? Below is the Cayley table: note that modular multiplication is just multiplication but with the rule 3=0 since we are in the modular world, hence 2 × 2 = 4 = 3 + 1 = 1.

 1 2 1 1 2 2 2 1

This is a commutative group since 1 is the identity element, 2 has order 2, and ab=ba for any elements in the table. We also note that, for any n, these operations will ‘play nice’ since we have a(b+c)=ab+ac in $\mathbb{Q}$ and so a(b+c) will either be a(b+c) or a(b+c-n) which when multiplied out will give a(-n)+ab+ac or ab+ac. Note that $\mathbb{F}_4$ is not a field since it is not closed under multiplication: 2 × 2 is 0 in the modular world.

But what has this got to do with groups? As promised, now for the examples. We may produce some new examples of groups by looking at matrices with the usual matrix multiplication. Matrix multiplication over a field will always be an associative operation. So we just need to check the other group axioms i.e. do our matrices

• include the identity matrix (as the identity)?
• include an inverse for each element?
• form a closed set (so that AB is in the set for all pairs of matrices A and B)?

The matrices we will consider will take their entries from the field $\mathbb{F}_3$, and all be of the form

$\begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix}$

where: 0 is the identity element of ($\mathbb{F}_3$, +); 1 is the identity element of ($\mathbb{F}_3$\{0}, ×); and the a, b and c can be any elements of the field $\mathbb{F}_3$. Before continuing: what does the multiplication look like here? Try multiplying two general elements and also compute the order of any group element. Use this to check that this is a group (with matrix multiplication).

The matrix with a, b, and c equal to 0 is our identity and, by computing powers of a general element, we see that no element has order 2 and every non identity element has order 3. Since a, b, and c can each take 3 different values, this group has 27 elements. In summary, this group (let’s call it G) has

• 27 elements
• with 26 of order 3 (and the other the identity element).

We will now construct another group of order 27 such that 26 of these elements have order 3. We want lots of elements of order 3, and so we’ll put together lots of C3‘s (cyclic groups of order 3). But how? Let’s use the direct product.

The structure of C× C3. It’s worth checking this for yourself, but from an earlier exercise we have that this is not the same as C9. Considering an arbitrary element (a,b) in this group,

(a,b)(a,b)=(a2, b2)

and so all of the non identity elements of this group will have order 3. Similarly the group H:=C3×C3×C3 will have elements of the form (a,b,c) and multiplication again is component-wise i.e.

(a,b,c)(a,b,c)=(a2, b2,c2)

so that again all of the non identity elements have order 3. Thus H has

• 27 elements
• with 26 of order 3 (and the other the identity element).

Look familiar? So we have two groups with the property we were looking for. Now we need to show that they are not isomorphic. Note that H is commutative. However G is not. For example

$\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} =\begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} \ne \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}= \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}.$

This means that there cannot be an isomorphism $\phi$ from H (the direct product) to G (the matrices). If there were, it would send

$a\mapsto\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix},\; b\mapsto \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}$

for some non trivial a and b in H. First we note that

$\phi(aba^{-1}b^{-1})=\phi(1)=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$

and then, since it is a homomorphism

$\phi(aba^{-1}b^{-1})=\phi(a)\phi(b)\phi(a^{-1})\phi(b^{-1})= \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}^{-1}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}^{-1}$

which is non trivial by our matrix manipulation above. Note that this idea can be extended to see that a commutative group and a non commutative group can never by isomorphic. Our working has also meant that, in order to decide that they were not isomorphic, we didn’t have to draw out the entire Cayley table for each of them; and with 27 elements, that wouldn’t have been fun (it would have 272=729 entries)!

Thanks for reading and see you soon!

Advertisements