Print preview
Handout Friday 11/14
-
Now we can say more. In fact, the kernel of any homomorphism is a normal subgroup of \(G_1\text{.}\)
-
A normal subgroup is one for which \(aH = Ha\) for all \(a \in G\text{.}\) Another way to say this: for all \(a \in G\) and \(h \in H\text{,}\) \(ah = h'a\) for some \(h' \in H\text{.}\) This is the same as saying \(aha^{-1}= h'\text{,}\) which is to say that \(H\) is closed under conjugates.
-
Now let’s prove that the kernel of a homomorphism is a normal subgroup.Suppose \(a \in K\) and consider any conjugate \(x a x^{-1}\) where \(x\) is some element in \(G_1\text{.}\) Now \(f(a) = e\text{.}\) What is \(f(x a x^{-1})\text{?}\) Well, by the homomorphism property,\begin{equation*} f(x a x^{-1}) = f(x) f(a) [f(x)]^{-1}= f(x) e [f(x)]^{-1}= e \end{equation*}So \(xax^{-1}\in K\text{.}\) Thus \(K\) is closed under conjugates, so is a normal subgroup.
-
This is huge! We like homomorphisms because they let us focus our attention on one property of a group. So it would be nice to know exactly what sorts of groups are homomorphic images of a given group.
What about rings? Of course we can defined a homomorphism for rings just like groups. Now we need both operations to satisfy the “isomorphism” property.
-
We can still consider the kernel of the homomorphism, and this is still the set of things that map to the group identity. (It is true that if there is a multiplicative identity, it must be sent to the multiplicative identity, but we don’t think of that as a kernel.)
-
In this context, we can say more about the kernel. Not only is it a subring of the domain, it is also an ideal! Let’s prove this.
-
Let \(K\) be the kernel of a ring homomorphism. Let \(a \in K\) and \(r \in R\text{.}\) We want to show that \(ra \in K\text{.}\) Well see where it goes!
-
We have \(\varphi(ra) = \varphi(r)\varphi(a) = \varphi(r)\cdot 0 = 0\text{.}\) So \(ra\) is sent to zero, so it is in the kernel. Tada.
