Checking that the given function is really a homomorphism.

Question

Here is the question and its answer:

(a) To what familiar group is $GL_{2}(\mathbb{Z}_2)$ isomorphic?

Answer.

$(a)$

$GL_2(\mathbb{Z}_2)$ is the set of invertible (non-zero determinant) $2\times 2$ matrices whose entries are in $\mathbb{Z_2}.$ Explicitly its elements are $$A= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},B= \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}, C= \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix},D = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, E= \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}, F= \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}.$$

The Cayley table for this group is as follows:

$$ \begin{array}{c|cccccc} \times & A & B & C & D & E & F \\ \hline A & A & B & C & D & E & F \\ B & B & C & A & F & D & E \\ C & C & A & B & E & F & D\\ D & D & E & F & A & B & C\\ E & E & F & D & C & A & B \\ F & F & D & E & B & C & A \end{array} $$

The Cayley table is not symmetric about the principle diagonal hence $GL_{2}(\mathbb{Z}_{2})$ is not commutative. there is only one non-commutative group of order 6 which is $S_{3}.$ Hence $GL_{2}(\mathbb{Z}_{2}) \cong S_{3}.$ And the isomorphism we can define as follows:

$() \rightarrow A, (12) \rightarrow D, (13) \rightarrow F, (23) \rightarrow E, (1 2 3) \rightarrow B$ and $(132) \rightarrow C$. It is easy to check that it is an isomorphism.

My question is:

How can I check that this function is really a homomorphism? do I have to check the 30 permutations or is there a smart way of checking that it really defines a homomorphism?

Answer 1

To define a homomorphism $GL_2(\mathbb{F}_2) \to S_3$ is the same thing as to give an action of $GL_2(\mathbb{F}_2)$ on three objects. What three objects does $GL_2(\mathbb{F}_2)$ act on? Exactly the three nonzero vectors $(1, 0), (0, 1), (1, 1)$ in $\mathbb{F}_2^2$! (In other words, the projective line $\mathbb{P}^1(\mathbb{F}_2)$.)

This gives you a homomorphism, automatically, and now you just need to check that it's injective and surjective. Injectivity is easy, e.g. the columns of a matrix are determined by how it acts on $(1, 0)$ and $(0, 1)$. Surjectivity can be proven in several different ways, which generalize differently to different situations:

1. It follows from injectivity once you've computed that both groups have order $6$.
2. It follows once you show that $GL_2(\mathbb{F}_2)$ has both an element of order $2$ and an element of order $3$.
3. It follows from the more general fact that $GL_n(F)$ acts transitively on the set of pairs of linearly independent vectors in $F^n$, for any field $F$ and any $n \ge 2$, and from the observation that any two nonzero vectors in $\mathbb{F}_2^2$ are linearly independent.

The same style of argument gives many other exceptional isomorphisms. For example, $PGL_2(\mathbb{F}_3)$ naturally acts on $4$ objects, namely the projective line $\mathbb{P}^1(\mathbb{F}_3)$, and this gives a homomorphism $PGL_2(\mathbb{F}_3) \to S_4$ which is an isomorphism and which restricts to an isomorphism $PSL_2(\mathbb{F}_3) \cong A_4$. (We can again argue by showing that injectivity follows in general and then computing that both groups have the same order, and we again have a more conceptual argument using transitivity.)

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Edit: There's also a similarly very clean way to define a homomorphism $S_3 \to GL_2(\mathbb{F}_2)$, as follows. By definition $S_3$ acts on a set $X$ with three elements. We want to canonically produce a $2$-dimensional vector space over $\mathbb{F}_2$ from $X$ somehow. We can do this by first taking the free vector space $\mathbb{F}_2[X] \cong \mathbb{F}_2^3$ (which $S_3$ acts on but which has dimension $3$), then restricting to the subspace

$$\{ (x, y, z) \in \mathbb{F}_2^3 : x + y + z = 0 \}$$

which $S_3$ also acts on and which now has dimension $2$. This gives us a map $S_3 \to GL_2(\mathbb{F}_2)$ (and $\mathbb{F}_2$ can be replaced by any field here) which turns out to be the inverse of the above map! Well, once we've picked a suitable basis of this subspace anyway. I learned this from these notes on a talk by Serre.

Answer 2

What you should do is not define the function (homomorphism) on all 6 elements. Then you have the burden of showing it is a homomorphism. As $S_3$ is generated by a transposition and a three-cycle together, send those two elements respectively to elements of order 2 and 3 respectively. Then define it on other elements to make it a homomorphism. that is a is sent to A and b is sent to B then define it on $a\circ b$ to be the matrix product $AB$.