In the previous post we recapped Cayley tables and did lots of comparing of the groups given in this form. We know that the Cayley table captures structure…so now we’ll define exactly what it means for groups to be the ‘same’. If two finite groups G and H are the ‘same’, they must be a relabelling of each other. Since the Cayley table tells us the structure, if we can see that the tables are the ‘same’ then we may say that the groups are the ‘same’ (up to permuting the rows and columns of the Cayley tables).

The proper word for same is *isomorphic*. We say that G is *isomorphic* to H and that there is an *isomorphism* between G and H. The aim for the rest of this post is to provide an algebraic condition for isomorphic. We do this for two reasons. First, it allows us to check more easily if two finite groups are isomorphic. Secondly, it extends the idea of isomorphic to infinite groups.

We want to re-evaluate what our condition on two groups having the same Cayley table means. A clear condition (to start with) is that there must be…

- a bijection between the elements of the two groups

which gives us that the two groups must be of the same order (contain the same number of elements). But it’s more than that, since our final example had two groups of order 4 which were not isomorphic: they were not the ‘same’. So what is it that we were checking in the previous post? We want our relabelling to preserve the inside of the Cayley table. Let’s have a picture (where G is on the left and H is on the right. This example generalises to bigger groups).

Here, is the bijection from a, b, c, d to the four elements of H. We now want to think about what it means for these Cayley tables to be the same. Since we want to be a relabelling, whatever ab turns out to be, when we look at (ab) it should match with whatever (a)(b) turns out to be. If this works everywhere i.e. for every pair x, y in G we have (xy)=(x)(y), then the Cayley tables will be of the same structure (just a relabelling). We now have:

- Two groups G and H are
*isomorphic*if there exists an*isomorphism*from G to H.

**Definition. **An *isomorphism* from G to H is a function which…

i) is a bijection from the elements of G to the elements of H

ii) satisfies, for all x, y in G: (xy)=(x)(y).

Once you’ve seen a definition, it’s worth trying some examples to get to grips with it.

First ideas (cyclic groups): consider, as we did in the previous post, the two ways we interpretted the cyclic group of order m. Look back if you can’t remember. What’s the most succinct way to describe an isomorphism between these groups i.e. what is the least number of statements of the form (x)=n that you can write down to determine the isomorphism? Also, how many isomorphisms are there? (It’s not enough to do just one example here!) Finally, can you find a copy of C_{m} in C_{2m}? Use this to make a conjecture about when you can find a copy of C_{m} in C_{n}.

Second ideas (orders of elements): imagine, as part of an isomorphism, we have (x)=a. If x has order m, can anything be said about the order of a? If x has infinite order, can a have finite order? Hence, what can we say about the orders of elements of G and H if they are isomorphic?

Third ideas (generators): consider if a group G is generated by elements x_{1}, …, x_{n} (if G is a cyclic group, then x_{1} can be sufficient to generate G). If : G H is a isomorphism, then do (x_{1}), (x_{2}), …, (x_{n}) generate H? This might require some thought. I’d have a think about the cyclic case and then try and think about what it means for elements to generate a group and where must send inverses of elements.

There are two parts to the isomorphism definition. What would happen if you dropped one of the conditions? If we dropped the second condition (which essentially means the map is structure preserving) then all that we’d have is that the map is a bijection: that the underlying sets of the groups are the same. This is not particularly interesting since there are many non-isomorphic groups of the same order. But what about dropping that the map need be a bijection? This has a name.

**Definition. **Let G and H be groups. A *homorphism* : G H is a function which satisfies, for all x, y in G: (xy)=(x)(y).

We’ll discuss these in the next post. Until then, try and come up with some homomorphisms which are not isomorphisms.