Homomorphisms are more general than isomorphisms, except that need not be a bijection. For example, suppose and . Then define by . This is *not* injective, because all go to . It’s a homomorphism.

**Lemma: ***Suppose is a homomorphism. Then **, **, and i**f , .*

*Proof: *First: If , then , so is its own inverse, and .

Second: , so .

Third: We want to prove that it’s closed under multiplication and inverses. Suppose . Then for some . Thus so . This shows that it’s closed under multiplication. Suppose . Then for some . Now so , i.e., it’s closed under inverses. QED.

**Definition: **Suppose . Then is *normal* if for all .

This implies that , and actually, . If is abelian, i.e., commutative, then any subgroup of is normal.

**Definition: **Let be a homomorphism. Then define . This is called the *kernel* of .

**Proposition: ***Let be a homomorphism. Then if and only if is injective.*

*Proof: *Suppose is injective. Then suppose (i.e. is a collection of elements of which is indexed by some set ). This means that . But is injective. Thus ; since is arbitrary, this means that .

Now suppose , and suppose . We want to prove that (because this is the *definition* of an injective function). If , then . So . Thus ; but , and so . QED.

**Corollary: ***Suppose is an isomorphism. Then .*

What was the point of introducing the notion of a normal subgroup before, though?

**Theorem: ***Let be a homomorphism. Then is a normal subgroup of .*

*Proof: *We first need to show that is a subgroup of . It suffices to prove that it is closed under multiplication and inverses. For multiplication: suppose . Then , so . For inverses: suppose . Then , so . So is a subgroup of .

Now to prove normality. Suppose . We need to show that if , then . Well, , which is equal to , so for any and . QED.

We can ask: if is a homomorphism, is isomorphic to a subgroup of ? In general, it can’t be the whole group itself, because the kernel of could be nontrivial, i.e., more than one element could map to the identity. (Unless the homomorphism is injective, because then is trivial.) It turns out that the subgroup of which is isomorphic to is something called the *quotient group* of . (The *thing* it’s isomorphic to is called the quotient group, but the *reason* it’s isomorphic to the quotient group is the content of Talk 8, namely the *isomorphism theorems*. These are the most important theorems in group theory.)

Another, perhaps more simpler way of motivating quotient groups is as follows. Suppose is a group, and . Can we find a group and a homomorphism such that is isomorphic to ? (Because is normal, you obviously need to be a normal subgroup of .) Defining the quotient group will be the topic of the next post.

Have fun,

SKD