A homomorphism between two groups does NOT have to be a one-to-one function. If it’s
not one-to-one, then there’s a group associated with the homomorphism which measures the degree
to which the function is not an injection. This group is called the kernel...
To really understand the kernel, there are a few properties you need to learn about homomorphisms:
homomorphisms send identities to identities, and inverses to inverses. We’ll begin by
proving these properties. Suppose we have two groups G and H, and a homomorphism F between
them. Recall that a function F is a homomorphism if F of (X-times-Y) equals (F-of-X) times
(F-of-Y). Don’t forget that the operation on the left hand side is for the group G,
while the operation on the right hand side is for the group H. These operations can be
different from one another.
We’ll first show why group homomorphisms send the identity in G to the identity in
H. So we don’t confuse the identities of these two groups with each other, let’s
denote the identity in G as 1-sub-G and the identity in H as 1-sub-H. We begin by picking
a random element in X that is different from the identity. Any element will do. From the
definition of the identity element, X times 1-sub-G equals X. Now look what happens if
we apply F to both sides of this equality. We get F-of-(X times 1-sub-G) equals F-of-X.
Since F is a homomorphism, we can write the left side as (F-of-X) times (F-of-1-sub-G).
Now F-of-X is an element in H; let’s call it Y. This gives us Y times (F-of-1-sub-G)
equals Y. And since H is a group, Y has an inverse. If we multiply both sides by the
inverse of Y we get F of 1-sub-G = 1-sub-H. This proves that homomorphisms send identity
elements to identity elements.
Next we’ll show why homomorphisms send inverses to inverses. Suppose F-of-X equals Y. Note
that X is in G and Y is in H. We want to show that F-of-X-inverse = Y-inverse. To see why,
we’ll use the fact that X times X-inverse equals 1-sub-G. If you apply F to both sides
you get F-of-(X times X-inverse) equals F-of-(1-sub-G). We just showed that F-of-(1-sub-G) = 1-sub-H.
And on the left, we can use the property of homomorphisms to get F-of-X times F-of-(X-inverse)
equals 1-sub-H... F-of-X equals Y, so we get Y times F-of-(X-inverse) equals 1-sub-H. Multiplying
both sides on the left by Y-inverse gives us F-of-(X-inverse) equals Y-inverse. This
proves that homomorphisms send inverses to inverses.
We now have all the tools we need to define and understand the kernel. Suppose that F
is not one-to-one. Then there are at least two elements in G which map to the same element
in H. To be concrete, suppose X1, X2, and so on are the elements which map to Y. This
means F-of-X1 equals Y, F-of-X2 equals Y, and so forth. Now watch what happens if we
multiply all of these equalities by F of X1-inverse... Since homomorphisms send inverses to inverses,
(F-of-X1-inverse) equals (Y-inverse). This allows us to simplify the right hand sides…
Next, we can combine the left hand sides because F is a homomorphism… Because F is not one-to-one,
we find that there are multiple elements in G which all map to the identity in H. These
elements are called the kernel of F and we write the kernel like this… This notation
emphasizes that the kernel is a property of the homomorphism, NOT the groups.
If I told you a homomorphism F was not an injection, it’s not obvious that there would
be more than one element that maps to the identity element. But there are! If F is not
one-to-one, then the kernel contains more than one element. Think of the kernel as a
way to measure the degree to which F fails to be one-to-one. We know for every homomorphism
the identity in G maps to the identity in H, so the kernel is never empty; it always
contains the identity 1-sub-G. And if the kernel only contains the identity, then F
is one-to-one.
The kernel is a subset of G, but it’s actually more than that. It’s also a SUBGROUP of
G. To test your understanding of homomorphisms and kernels, I’d like you to check that
the kernel is, in fact, a subgroup. If you get stuck, you can always ask a question in
the comment box below.