## Homomorphisms

[ This post roughly corresponds to Chapter VI of the old blog. ]

For sets, one considers functions fS → T between them. For groups, one would like to consider only actions which respect the group operation.

Definition.  Let G and H be groups. A function f : G → H is a homomorphism if f(xy) = f(x)f(y) for all x, y in G. Note that the product on LHS is in group G, while the product on RHS is in group H.

A homomorphism is called a monomorphism (resp. epimorphismisomorphism) if it is injective (resp. surjective, bijective). Clearly, if fG → H and gH → K are both homomorphisms of groups, then so is gfG → K. Furthermore, f satisfies the following.

• $f(e_G) = e_H$ : we have $f(e_G) = f(e_G * e_G) = f(e_G) * f(e_G)$, so cancel $f(e_G)$ on both sides;
• $f(x)^{-1} = f(x^{-1})$: we have $f(x)f(x^{-1}) = f(x*x^{-1}) = f(e_G) = e_H$, so multiply both sides by $f(x)^{-1}$.

These are the least we should expect out of homomorphisms. It also turns out a group homomorphism preserves subgroups, in the following sense:

Theorem.  Let f : G → H be a group homomorphism.

• If K < G is a subgroup then f(K) < H is a subgroup.
• If L < H is a subgroup, then f-1(L) < G is a subgroup.
• If L is a normal subgroup of H, then f-1(L) is a normal subgroup of G.

Definition. In particular, the image im(f) := f(G) is a subgroup of H, and the kernel ker(f) := f-1({e}) is a normal subgroup of G.

The proof is easy so we’ll leave it to the reader. Notice one missing case: i.e. if K is a normal subgroup of G, is f(K) normal in H? A moment’s thought tells us this is obviously not true: pick a subgroup H of G which is not normal; then the inclusion H → G has image precisely H, which is not a normal subgroup of G by construction.

This will be a recurring theme throughout algebra: pullbacks $f^{-1}(Y)$ are generally more well-behaved than images f(X).

## Examples

1. If N is a normal subgroup of G, then the projection map fG → G/N which takes g to gN is an epimorphism. The kernel is precisely N. In fact, this characterises all epimorphisms from G as we shall see from the isomorphism theorems.
2. If GSn (n>1), the set of n-degree permutations, then the sign map $f: G \to \{+1, -1\}$ which takes even (resp. odd) permutations to +1 (resp. -1) is an epimorphism. The kernel is An.
3. If G = GLn(R) denotes the group of n × n invertible matrices with real entries, then the determinant map det : G → R* is a group homomorphism. The kernel is SLn(R), the set of matrices with determinant = 1. ## First Isomorphism Theorem

Although there’re three, it’s really the first that’s most important. The remaining two can be derived from this first one.

First Theorem of Isomorphism. Let f : G → H be a group homomorphism with kernel N = ker(f). This gives an isomorphism φ : G/N → im(f), where φ(gN) = f(g).

Proof. There’re several things to check for the map φ.

• Well-defined & injective: $xN = yN \iff y^{-1}x \in N \iff f(y^{-1}x) = e \iff f(y)^{-1}f(x) = e \iff f(x) = f(y)$.
• Surjective: follows from definition of im(f).
• Homomorphism: $\phi(xN * yN) = \phi(xyN) = f(xy) = f(x)f(y) = \phi(xN)\phi(yN)$.

Proof complete. ♦

Pictorially, we get: Here’s one application of the theorem. Suppose HG is a subgroup and N is a normal subgroup of G. This gives homomorphisms H → G → G/N so the composed map is also a homomorphism. The kernel of the resulting map is $\{x\in H: xN = N\} = H\cap N$. So by the first isomorphism theorem, we get an injective map $H/(N\cap H) \hookrightarrow G/N$.

Here’s another example: let’s prove that

1. $GL_n(\mathbf{R})/SL_n(\mathbf{R})$ is isomorphic to the group R*;
2. if G denotes the set of complex numbers z with |z|=1, prove that G is isomorphic to R/Z, where Z=group of integers, R=group of real numbers, both under addition.

The method in both cases is identical. Construct a map from one group to the other; find the kernel, then apply the above first isomorphism theorem.

1. Take the determinant map $GL_n(\mathbf{R}) \to \mathbf{R}^*$. The map is surjective (e.g. the diagonal matrix with one entry x and remaining entries 1 has determinant x). The kernel is $SL_n(\mathbf{R})$. So the result follows from the first theorem of isomorphism.
2. Take the exponential map R → C*, where x goes to exp(2πix). The kernel is exactly the set of integers while the image is the set of complex numbers z with |z|=1. The result follows from the theorem again.

“Factoring Through”

Let fG → H be a homomorphism. Suppose we know of elements xy for which f(x) = f(y) = e. Then since the kernel of f contains both x and y, it must contain N(xy), the normal subgroup generated by {xy}. So we get a series of group homomorphisms: $G\to G/N(x,y) \to G/\text{ker}(f) \to \text{im}(f) \hookrightarrow H$,

where the first two maps are projections (since N(x, y) ≤ ker(f)), the third is due to the first isomorphism theorem and the fourth is just good old inclusion. We thus conclude that f gives rise to G/N(xy) → H.

This often happens when one doesn’t quite know the kernel,  but can obtain a few explicit elements in it. The question is whether these elements generate the full kernel. One can, at the very least, say that the map f factors through G/N(xy) → H.

## Second and Third Theorems of Isomorphism

Let’s look at the remaining two theorems of isomorphisms.

Second Theorem of Isomorphism. Let H, K ≤ G be subgroups with H normal in G. Then we already know KH is a group. We get the isomorphism $K/(K\cap H) \cong KH/H$.

Proof.

Map K → KH by inclusion and KH → KH/H by projection (this is well-defined since H is normal in KH). The composition is surjective because every element of KH/H is of the form (kh)HkH. The kernel is the set of k in K such that kHH, i.e. kernel = K ∩ H. Thus the result follows from the first theorem of isomorphism. ♦

Third Theorem of Isomorphism. Let N ≤ H ≤ G be subgroups such that N and H are normal in G (this also means N is normal in H). Then we have the isomorphism $(G/N)/(H/N) \cong G/H$.

Proof.

Construct a map fG/N → G/H via f(gN) = gH. This is well-defined since $gN = g'N \iff g^{-1}g' \in N \implies g^{-1}g' \in H \iff gH = g'H$.

It is clearly surjective. The kernel is the set of xN such that xHH, i.e. the kernel is exactly H/N. Now apply the first theorem of isomorphism. ♦ ## Correspondence between G and G/N

Let $N \triangleleft G$. The key result here is the interplay between subgroups of G and subgroups of G/N.

Theorem. There is a one-to-one correspondence : $\{H : N \le H \le G \text{ subgroup}\} \leftrightarrow \{K : K \le G/N \text{ subgroup}\}$

which preserves:

• inclusion : $H_1 \subseteq H_2 \iff K_1 \subseteq K_2$;
• normality : $H_1 \triangleleft H_2 \iff K_1 \triangleleft K_2$;
• indices : $[H_2 : H_1] = [K_2 : K_1]$ whenever $H_1\subseteq H_2$.

Here, $H_1$ and $H_2$ are subgroups on the LHS corresponding to subgroups $K_1$ and $K_2$ on the RHS.

Sketch of Proof.

Let $\pi : G \to G/N$ be the projection map. Let’s construct maps between the two sides.

• RHS → LHS : for subgroup K ≤ G/N, take $H = \pi^{-1}(K)$ which is a subgroup of G containing N = ker(π);
• LHS → RHS : for N ≤ H ≤ G, take $K = \pi(H) \le G/N$.

We’ll leave it to the reader to show that the composition of the two maps – in either order – is the identity map (on either the LHS or RHS). This shows the bijection.

• Inclusion follows immediately from the definition of the two maps.
• For normality, prove that K is normal in G/N iff $\pi^{-1}(K)$ is normal in G.
• For indices, prove that a complete set of coset representatives of $[\pi^{-1}(K_2) : \pi^{-1}(K_1)]$ is exactly a complete set of coset representatives of $[K_2 : K_1]$. ♦

## Example of a “Universal” Proof

Of course there’re more properties we haven’t mentioned: e.g.

If $H_i \leftrightarrow K_i$ is a collection of correspondences, then the intersection also corresponds: $\cap_i H_i \leftrightarrow \cap_i K_i$.

This can be proven in a straightforward way, but it’s “better” to use the existing properties. Now if $H = \cap_i H_i$, then H satisfies the property:

• $H \subseteq H_i$ for each i;
• if there is an H’ such that $H' \subseteq H_i$ for each i, then $H' \subseteq H$.

The corresponding K must then satisfy the same properties, albeit with Ki‘s:

• $K \subseteq K_i$ for each i;
• if there is a K’ such that $K' \subseteq K_i$ for each i, then $K'\subseteq K$.

Thus the only possibility for K must be the intersection of the Ki‘s.

[ Note: this proof is “better” in the sense that the approach is more high-level (technical term: it is more universal). While the universal method doesn’t seem any shorter now, it lends itself to replication for many different types of algebraic structures, some of whose definitions are extremely intricate and involved. ]

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