Explanations to Myself: The Isomorphism Theorems

There are four isomorphism theorems. The three they tell you are the isomorphism theorems and one called the lattice or correspondence theorem.

A homomorphism of a group is a wonderful thing: It’s a operation that tries to preserve as many properties about an object as it can when moving between two algebraic structures. Say f is a homorphism such that f: G → H, where G and H are groups. Then f is the function that tries to grab all the angles, lengths, etc that it can and ensure that those things in G stay the same under its transformation to the new group H. If it does than we say those objects are preserved under homomorphism or invariant under homomorphism.

A kernel on the other hand is a measure of how much a homomorphism fails to be injective. An isomorphism is an bijective map between two sets, its there and back again. Its a homomorphism that has achieved the ultimate goal of preserving everything it could get its hands on in the object that we would want to be preserved, no matter which set we start from. Two objects are isomorphic if you can’t tell them apart at least in terms of what you wanted or could preserve.

Theorem 1: Let f: G → H be a group isomorphism, where G and H are groups. Then,

  • The kernel of an isomorphism is a normal subgroup of G.
  • The image of the kernel is a subgroup of H,
  • The image of f is isomorphic to group G/(ker f).
  • If f is surjective, then the image of H is also isomorphic to H/(ker f).

This theorem basically says:

If G is isomorphic to H, then H is isomorphic to G.

This theorem actually stems from facts in category theory. Namely, that the morphism f has objects that can be deduced from it and that each of these groups has a kernel. Technically speaking that the normal epimorphism and monomorphisms for the category of groups form a factorization system for the category.

Second Theorem: Let G be a group, S any subgroup of G, and H a normal subgroup of G. Then,

  • The product SH is a subgroup of G.
  • The intersection of S and H is a normal subgroup of S,
  • And the quotient groups SH/H and S/(S⋂H) are isomorphic to one another.

In other words, the isomorphism preserves the order of the elements in the group. It isn’t actually necessary for H to be normal as long as S is a normalizer in H. If that happens then the intersection of S and H is not normal in G, but it is in S.

Third Theorem: Let G be a group and H a normal subgroup of G. Then,

  • If S is a subgroup of G so that H⊆S⊆G then S/H is a subgroup of G/H.
  • Every subgroup G/H looks like S/H, when H⊆S⊆G.
  • If S is normal in G and H⊆S⊆G, then S/H is normal in G/H.
  • Every normal subgroup in G/H looks like S/H when S is normal and H⊆S⊆G.
  • If S is normal in G where H⊆S⊆G, then (G/H)/(S/H) is isomorphic to (G/S). (Oh look mom they cancel)!

This longwinded irritant says: Any finite group G is isomorphic to a group of permutations. Yup thats it.

Fourth Isomorphism Theorem: Let G be a group H a normal subgroup of H, G’ be the set of all subgroups A of G where H⊆A⊆G, and H’ be the set of all subgroups of G/H. Then,

  • There exists a bijective map f: G’ → H such that f(A) = A/H, ∀A∈ G.
  • If A and B are in G, and A’ = A/H and B’ = B/H, then
    • A ⊆ B ⇔ A’ ⊆ B’
    • If A ⊆B then the index [B’: A’] = [B: A]
    • <A, B>/H = <A’, B’> where <A, B> is the subgroup generated by A union B.
    • A is normal in G if and only if A’ is normal in G/H.

This theorem says when H is normal in G, then there exists a bijection (the thing we wanted to begin with) from set set of all subgroups A in G such that A contains H. And that this bijection maps onto the quotient group G/H.

So the structure of the subgroups in G/H is exactly the same as the structure of the subgroups in G which contain H, where H is the identity element.

So nearly all properties of subgroups are preserved in their images under this bijection onto the subgroups of a quotient group. Which could have been said much more clearly than all that jargon we just translated.

Advertisements

Leave a Reply

Please log in using one of these methods to post your comment:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s