We have claimed that the key property that a bijection between two groups must have for it to be an isomorphism should be that \(\varphi(ab) = \varphi(a)\varphi(b)\text{.}\) For rings, we must “respect” both operations, so \(\varphi(a+b) = \varphi(a)+\varphi(b)\) and \(\varphi(ab) = \varphi(a)\varphi(b)\text{.}\)
Last week we used functions between groups to say what it means for two groups to be basically the same. These functions were bijections that satisfied the isomorphism property: \(\varphi(a)\varphi(b) = \varphi(ab)\text{.}\) What would happen if we didn’t require the function to be a bijection? We would have a homomorphism.
Consider the group \(\mathbb Z_{10}\text{.}\) How does this group relate to \(\mathbb Z_5\text{?}\) Note that \(\mathbb Z_5\) is not a subgroup, and since it doesn’t have 10 elements, these groups are not isomorphic. However, we can define a function from \(\mathbb Z_{10}\) onto \(\mathbb Z_5\text{.}\) In fact, we can do so and still satisfy the isomorphism property (henceforth called the homorphism property).
Check that this satisfies the homomorphism property (for a few cases at least).
Notice that if \(f:G \to H\) is a homomorphism then multiple elements in \(G\) can be sent to the identity in \(H\) (unless \(f\) is actually an isomorphism, which you will show on your homework). We do know that the identity in \(G\) will be one of them though.
To prove this, consider \(f(e) = f(ee) = f(e)f(e)\text{.}\) Whatever \(f(e)\) is, it has an inverse. Multiplying by the inverse gives us \(e_H = f(e_G)\text{.}\)
Let \(f:G \to H\) be a homomorphism. The kernel of \(f\) is the set \(K\) of all element of \(G\) which are carried by \(f\) onto the identity element of \(H\text{.}\) That is,
\begin{equation*}
K = \{x \in G ~:~f(x) = e\}
\end{equation*}
What sort of set is the kernel? Well, from the above facts it must contain the identity. What if \(a\) and \(b\) are both in the kernel? Then \(f(a) = e\) and \(f(b) = e\text{.}\) What about \(ab\text{?}\) Well \(f(ab) = f(a)f(b) = ee = e\) so \(ab\) must then also be in the kernel. And \(f(a^{-1}) = [f(a)]^{-1}= e^{-1}= e\) so \(a^{-1}\in K\) as well.
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.
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.)
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!
