These are some of the most important theorems in group theory. In this post, we’ll provide proofs, and explain some consequences of these theorems.
Theorem (first isomorphism theorem): Suppose is a homomorphism of groups. Then: is a normal subgroup of , is a subgroup of , and there is an isomorphism . If is a surjection, then .
The proof is uses a lot of concepts, so tread carefully.
Proof: We’ve already proven the first two statements. We’ll now prove the last one. Define the map as . This is well defined, as we saw earlier. It’s a homomorphism. We want to prove that it’s bijective, i.e., surjective and injective. It’s obviously surjective by definition.
We need to prove that it’s injective, i.e., if , then . Recall that this is equivalent to proving that the kernel is trivial, i.e., all we need to prove is that if , then . Well, if , then . So ; but this implies that , so it is indeed injective! Thus it’s a bijective homomorphism, i.e., an isomorphism!
If the map is surjective, then , so . QED.
Here is a cool application. Recall that denotes the set of invertible matrices of real numbers. Recall that for a matrix you can define the determinant function, which takes to . Analogously, the determinant of a matrix is . All of this can be generalized to matrices. What you should notice is that the determinant of an invertible matrix of real numbers is never zero, and is always a real number.
The group of all invertible matrices whose determinant is is defined to be . This is a subgroup of . Now, the determinant function is a homomorphism , where denotes the real numbers with the number zero removed, where the operation is multiplication. Well, the identity is , so the kernel of this homomorphism consists of those invertible matrices whose determinant is , and this is nothing but ! So by the first isomorphism theorem.
Now onto the second isomorphism theorem.
Theorem (second isomorphism theorem): Suppose , and let be a normal subgroup of . Define . Then , is a normal subgroup, and there is an isomorphism: .
Proof: For the first part: all we need to prove that it’s closed under multiplication and inverses. For multiplication: Suppose . Then and . So . Because is normal, you have for some . So by multiplication on the left. So , and this is an element of because and . For inverses: suppose . Then . We want to show that . Well, . For convenience, write and . Then because is normal, for some , and thus . So , and it’s therefore in .
For the second part: we want to prove that if , then . Suppose . Then is in , because and and is a group, i.e., is closed under multiplication. Because is normal, and as well, by definition, . Thus for any , we have and , i.e., .
For the last part: consider the map defined by . Then you can compose with the map , which takes . So the total map takes . Denote this composition by . OK, so now let’s consider some random . Then , so what we observe is that is surjective. Now, let’s ask: what is ? Well, , which is . But this is the definition of . So , which is a group because the intersection of subgroups is itself a subgroup. Now we use the first isomorphism theorem, which says that is isomorphic to . QED.
The very last isomorphism theorem is as follows.
Theorem (third isomorphism theorem): Suppose are normal subgroups. Then is isomorphic to .
Proof: Recall the natural map defined by . The kernel is , and it contains . Define by . Obviously this map is surjective. Now, its kernel is because if , then and . So consists of cosets for . But this is just . Now the first isomorphism theorem tells us that . But , so . QED.
This is kind of like cancellation. For example, this tells us that if . I guess we’ll end here for now; next time, we’ll go on to the Sylow theorems in group theory.