More on homomorphisms

Recall that a homomorphism \phi: G\rightarrow H was a structure preserving function that need not be injective or surjective (unlike an isomorphism). Structure preserving is completely captured algebraically by the identity \phi(x)\phi(y)=\phi(xy) holding for all x, y \in G.

First observations.

Let \phi: G\rightarrow H be a homomorphism, and e_G and e_H be the identity elements of G and H respectively. Then

    1. \phi(e_G)=e_H; and
    2. \phi(x^{-1})=\phi(x)^{-1} for all x \in G.

These are both standard results. Before continuing: have a go at proving these.

What do we know of e_G? The property that uniquely defines it is that e_Gx=xe_G=x for every x \in G. Thus \phi(x)=\phi(e_Gx)=\phi(e_G)\phi(x) and so \phi(e_G) is an element such that, for every x\in G\phi(e_G)\phi(x)=\phi(x). We could proceed in several ways, but here are two.

First way: \phi(e_G)\phi(x)=\phi(x)\Rightarrow\phi(e_G)\phi(x)\phi(x)^{-1}=\phi(x)\phi(x)^{-1} \Rightarrow \phi(e_G)e_H=e_H and so \phi(e_G)=e_H.

Second way: observe that, since it holds for all x\in G, we may set x:=e_G. Then \phi(e_G)\phi(e_G)=\phi(e_G). From earlier, the only idempotent (an element such that e^2=e) in a group is its unique identity element. Hence \phi(e_G)=e_H.

For the other proof, note that for any x\in G

e_H=\phi(e_G)=\phi(xx^{-1})=\phi(x)\phi(x^{-1}) and e_H=\phi(e_G)=\phi(x^{-1}x)=\phi(x^{-1})\phi(x)

so that \phi(x^{-1}) is the (unique) inverse of \phi(x).

Exercise: use these results to show, given any x, y \in G such that xyx^{-1}y^{-1}=e_G we have that \phi(x)\phi(y)\phi(x)^{-1}\phi(y)^{-1}=e_H. Then consider what you’ve shown (hint: the first equality is equivalent to xy=yx).

Of course, the above also applies to isomorphisms, since these are homomorphisms with the additional property that they are injective and surjective. Let’s consider these two definitions. Another word for injective is one-to-one. I think this really captures the definition. It means that each element in G is sent to one element in H. But it needn’t be surjective. A surjective map is one that ‘hits’ every point in H (every point in H has at least one preimage in G). Try and come up with an injective function \phi from \mathbb{Z} to \mathbb{Z} that is not surjective. Using addition with \mathbb{Z}, is your map a homomorphism? Can you find one?

Definition a monomorphism is an injective homomorphism.

Note that mono means one (in the same way as in sound systems: it’s the opposite to stereo sound) and so monomorphism meaning a one-to-one homomorphism makes sense. Also an injection can be thought of as an embedding: if G injects into H then every point of G is put somewhere different within H. Thus H can ‘cover’ G.

You can also have surjective functions that are not injective, for example from \mathbb{Z} to \mathbb{N}. There’s a reason I haven’t asked you to find a surjective homomorphism from \mathbb{Z} to \mathbb{Z} that isn’t injective, which we’ll see in a bit. In the same way as injections are like embeddings, a surjection can be thought of as a covering. If G surjects onto H, then G can ‘cover’ every point in H. You can visualise this as G ‘covering up’ H. Note that this is the dual (opposite) to being injective. These visualisations fit with the standard notation: if \phi is injective we write \phi: G\hookrightarrow H and if it is surjective we write \phi: G\twoheadrightarrow H.

Definition an epimorphism is a surjective homomorphism.

These ideas hint at a definition that we will not use, but I include for interest.

Definition Let G be a group. Then it is Hopfian if every surjective homomorphism from G to G is injective. It is co-Hopfian if every injective homomorphism from G to G is surjective.

This is quite a difficult definition. But in order to understand it, let’s have a play.

Example 1. If G is finite, then it is Hopfian and co-Hopfian. Why? This comes from a broader fact. If X is a finite set and \phi is a function from X to X, then \phi injective implies that \phi is surjective. Moreover \phi surjective implies that \phi injective. Since a homomorphism is a function with an extra property (structure preserving) we obtain the result.

Example 2. \mathbb{Z} is Hopfian but not co-Hopfian. The second statement follows from the exercise above: we can find an injective homomorphism that is not surjective. Think about this for a bit. Spoiler: this function is a monomorphism from \mathbb{Z} to \mathbb{Z}. The fact that \mathbb{Z} is not co-Hopfian is the reason I didn’t ask you to find a surjective homomorphism from \mathbb{Z} to \mathbb{Z} that wasn’t injective: there isn’t one!

I think I’ll give some more examples and proofs in a later post, since these definitions are more of a distraction from what we’re really looking at.

It’s worth noting at this point that if \phi: G\rightarrow H is a homomorphism, then \phi(G) needn’t be isomorphic to a subgroup of G. In order to justify this we need just one example. \mathbb{Z} is a nice group, so let’s play with that. What do subgroups of \mathbb{Z} look like? Are they all cyclic? Before continuing: find a homomorphism from \mathbb{Z} to a finite group.

So what’s the answer? Imagine if \phi: \mathbb{Z}\rightarrow G and \phi(3)=e, the identity element of G. Since \phi is a homomorphism, we have (for every n\in \mathbb{Z}) that \phi(3n)=\phi(3)^n=e. Moreover

\phi(3n+1)=\phi(3)^n\phi(1)=\phi(1);

\phi(3n+2)=\phi(3)^n\phi(2)=\phi(2);

and \phi(1)\phi(2)=\phi(3)=e.

So if \phi(1)\ne e, then we have that \phi(\mathbb{Z}) consists of 3 elements: e, \phi(1), \phi(2). What’s the group? It’s cyclic. It has order 3. Hence it is C_3. But be careful! We don’t know what G is. For example G could be C_3 \times H for any group H, since then

\phi(1)=(1, e_H), \phi(2)=(2, e_H), and \phi(3)=(0, e_H)=e.

Can you see how general the above example is? Which other cyclic groups does \mathbb{Z} surject onto?

In the next post we’ll consider whether there is an epimorphism from S_n to C_2.

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More on homomorphisms

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