Today we’ll be taking off from before and talk about isomorphisms.

Let and be groups such that there is a map of the underlying sets .

**Definition: **The map of the underlying sets is an *isomorphism* if is a bijection (i.e. one-to-one, i.e., the following two conditions are satisfied: 1) if in then in , and 2) for every , there is some such that . Alternatively, the map should have an inverse.) and respects the group structure, i.e., if is the operation of and is the operation of , then . If there’s a map between groups which is an isomorphism, we say that and are *isomorphic*, and we write .

Remember that we usually drop the operation’s symbol, so can be written as . What’s one condition for two groups to be isomorphic?

**Proposition: **If and are cyclic, and , then .

*Proof: *Since and are cyclic, write and . Then define a map which respects the multiplication as , so . The map obviously has an inverse, namely the map sending to . We only need to check that . Since every is of the form , we see that .

Another slightly more concrete isomorphism is the map defined by . Now suppose is a general group. An isomorphism is called an *automorphism*. For example, the map defined by is an automorphism. To see this it suffices to note that the inverse is , because obviously respects the group structure.

Let’s talk about something combinatorial. Let be a set. A *permutation* of is a bijection . For example, if is the set , the identity defined by is a permutation, as is the map for and . Alright, so suppose we’ve got two permutations . Then the composition is a permutation. This is true because its inverse is just . Also, if is a permutation, then so is its inverse . Let denote the set of permutations of .

**Lemma: ** is a group with the operation of composition.

*Proof: *We already stated that it’s closed under composition and inverses. The identity is, well, the identity function. Composition of functions is associative, so we’re done!

Now we’ll specialize to the case where . We claim that has elements. We’ll prove this. If , then there are possibilities for which element of we can map to. Now let such that . Because the maps in consideration are bijections, can’t be mapped to the same element which is mapped to. So there are possibilities for which element of can be mapped to. Doing this for each element of gives us possibilities for the permutations . Well, , so we’re done!

If , we write as and call it the th *symmetric group*. This actually isn’t abelian for . Alright, so let’s take a look at . This has elements. Now this is a group of *prime order*, and so is cyclic. What about ? This is *also* cyclic, and has order . So we see, using the above result, that there’s an isomorphism . (In fact, because for higher values of , isn’t of prime order (because isn’t prime), it’s not cyclic for .) There’s a very interesting (but not really helpful) result on *general* groups relating them to permutation groups:

**Cayley’s theorem: **Let be a group. Then is isomorphic to a subgroup of a permutation group.

*Proof: *Let be the map which takes to the following permutation of : . We note that is an isomorphism because it respects the multiplication and has an inverse given by .

Let denote the image of . This is a group because is a group. We want to show that is isomorphic to . We just need to check that the map is a bijection, because, as you can check, it respects the group structure (i.e. the multiplication). Recall that this means the following: 1) if in then in , and 2) for every , there is some such that .

First we check the first condition. Suppose in . Then let denote the maps , respectively. We want to show that . Well, what does this mean? It means that for any , . But this is automatically true if . Ok, now it is incredibly easy to see that for every , there is some such that , because *by definition* satisfies this! So is isomorphic to , which is in turn a subgroup of the permutation group .

**Corollary: **If is finite, then is isomorphic to a subgroup of .

*Proof: *Trivial.

Here’s a cool isomorphism. Let denote the group of symmetries of a regular -gon. For example, is simply the group of symmetries of an equilateral triangle. Drawing it out tells us that there are three rotation operations (namely: do nothing (the identity), rotate by , and rotate by ) and three flipping operations (namely flipping around the line drawn through one of the vertices which bisects the opposite edge). In total, . Wait a second! We also know that . Is it possible that ? Yes! They are, indeed, isomorphic! This makes sense, obviously, because the elements of permute the vertices of the equilateral triangle, which is *exactly* what does!

We may spend more time on the symmetric group later, but we should get the notion of a *homomorphism* out of the way first, which generalizes the notion of an isomorphism. The map defined by taking will be an example of a homomorphism (but note that it’s *not* an isomorphism!). That will be something for next time.

Have fun,

SKD