Name: The Kernel of a Group Homomorphism – Abstract Algebra
Uploaded: 2020-03-06T16:36:44.000Z
Duration: 4 min 53 s
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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

Basic passive voice

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

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

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

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

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The Kernel of a Group Homomorphism – Abstract Algebra

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