A homomorphism is a way to compare two groups for structural similarities. It’s a function
between two groups which preserves the group structure in each group. Let’s be more specific.
Suppose we have two groups G and H. Remember, G and H are different groups, and they may
have entirely different group operations. To help us keep G and H separate in our minds,
let’s use an asterisk for the group operation in G and a diamond for the group operation
in H. We’ll still use the word “times,” but we’ll write the correct symbol for the
operation.
Now pick any two elements X and Y in G. Let’s say that X times Y equals Z. Next, suppose
we have a function F that maps G to H. The three elements X, Y and Z are mapped to elements
in H; X is mapped to F-of-X, Y is mapped to F-of-Y, and Z is mapped to F-of-Z. The whole
point of a homomorphism is to find a structural similarity between two groups. So if X times
Y equals Z in G, then we’d like F-of-X times F-of-Y to be equal to F-of-Z in the group
H. So (F-of-X) times (F-of-Y) equals (F-of-Z). But Z equals X times Y. Substituting this
gives us (F-of-X) times (F-of-Y) = F-of-(X-times-Y). This is the definition of a homomorphism.
It’s a simple definition, but it captures a big idea - a way to compare two groups.
Let’s see an example. Let G be the group of real numbers under addition, and H be the
positive real numbers under multiplication. G is an abelian group with identity element
0, and H is an abelian group with identity element 1. Here’s the homomorphism. Define
the function F that maps X to (e-to-the-X). To make sure this is a homomorphism, we need
to verify that F-of-(X-plus-Y) = (F-of-X) times (F-of-Y). Don’t forget that the group
operation for G is addition, while the group operation for H is multiplication. By the
definition of F, this says that e to the (x plus y) equals (e-to-the-x) times (e-to-the-y).
And this is true! This is one of the rules of exponents. So F is a homomorphism.
Here’s another example of a homomorphism. G will be the real numbers under addition,
and H will be the complex numbers with absolute value 1. These complex numbers form a group
under multiplication. Showing this is a good exercise, so why don’t you pause the video
and show that H is a group. Here’s a hint - every complex number with absolute value
1 can be written as e to the i-THETA for some angle theta.
Define the function F as F-of-X equals e-to-the-i-x. For F to be a homomorphism, we must have F-of-(X-plus-Y)
= F-of-X times F-of-Y . This is true only if e-to-the- i-times-(x plus y) equals (e-to-the-i-x)
times (e-to-the-i-y). We can see this is true by expanding the left hand side… then using
the rules of exponents. It’s true; F is a homomorphism. By the way, I’d like to
point out that F-of-zero equals F-of-(2 pi). Both are equal to 1. In fact F of ANY multiple
of 2 pi is 1. So this is NOT a one-to-one function. Homomorphisms do NOT have to be
one-to-one.
Take a closer look at the word “homomorphism.” It’s a combination of “homo” which means
“same” and “morph” which means shape. But don’t be misled by this name. If two
groups are homomorphic, it doesn’t mean they’re the same. It just means they’re
SIMILAR. Also, not every pair of groups will have a non-trivial homomorphism between them. Some groups
are so different from one another there are no structural similarities at all.
One final note. How would you simplify this expression?
Well, the square root of negative one is i,
so this leaves us with “i subscribe”. {mouths the phrase:: “I’m so sorry…”}