# Abstract algebra, show that phi is a homomorphism

1. Nov 23, 2013

### usn7564

1. The problem statement, all variables and given/known data

3. The attempt at a solution

I'm very new to this kind of maths, so don't quite know how to get started. If I understood the question at all we have

$$g_i \mapsto \phi_i$$

and so I have a homomorphism if I can show that

$$\pi(g \cdot g_i) = \pi(g) \circ \pi(g_i)$$

I'm thinking it's trivially injective (might be way off here) because each g maps to a unique element in the symmetric group so there's not much to show.

But to show the homomorphism? Frankly don't have a clue
$$j = \phi_g (i)$$
$$\pi(g \cdot g_i) = \pi(g_j) = \phi_{g_j}$$

$$\pi(g) \circ \pi(g_i) = \phi_g \circ \phi_{g_i}$$

and then I come to a halt. How do I approach this?

2. Nov 23, 2013

### Mandelbroth

It is trivially injective, so I'll let you show that in all of its glory if you really want to.

If it's alright with you, I'm going to drop the multiplication signs and just juxtapose two elements to multiply them.

We wish to show that $\phi_{gg_i}=\phi_g\phi_{g_i}=\phi_{g_j}$ for $g,g_i,g_j\in G$. Do you see how to continue?

3. Nov 23, 2013

### usn7564

Not quite no, I mean I realize I have to find the connection between the expressions but don't know how to go about it. I'm assuming the j is a key here but as far as I can tell the permutation is more or less arbitrary so think I'm missing what information is entailed in the j.

4. Nov 23, 2013

### Dick

It can be a little awkward to express. Let's write down indices for everything we will need first. Suppose $g_i$, $g_j$ and $g_k$ are any elements of your group. Then $g_i g_j g_k=g_m$ for some m. And $g_j g_k=g_n$ for some n. So then we also have $g_i g_n=g_m$. Now you want to show $\phi_{g_i g_j}(k)=\phi_{g_i}( \phi_{g_j}(k))$. Work out both sides. Start by telling me what $\phi_{g_i g_j}(k)$ is in terms of the indices we've defined. And showing it's injective is easy, but you do have to say why.

Last edited: Nov 23, 2013