# Group theory question

1. Nov 29, 2007

### ehrenfest

1. The problem statement, all variables and given/known data
If G is a group, is it true that right multiplication by a given element is a homomorphism but left multiplication is not? That does not really make sense to me because aren't right and left multiplication "symmetric"?

2. Relevant equations

3. The attempt at a solution

2. Nov 29, 2007

### morphism

In general neither is a homomorphism.

3. Nov 29, 2007

### Office_Shredder

Staff Emeritus
Note to be a homomorphism, it needs to map the identity to the identity. Only a select few elements can multiply with the identity and give the identity back

4. Nov 29, 2007

### ehrenfest

I think I asked the wrong question.
Let G = {g_1, ..., g_n} be a group. Let X be a graph with vertices {g_1, ..., g_n}. Let $$\alpha_{g}$$ be the permutation of the vertices defined by $$\alpha_{g}(g_i) = g_i*g$$

Given a bijection between the elements of G and the elements of the group of \alpha_g, I want to show that the group of \alpha_{g} and G are isomorphic.

So, let $$h(g_i) = \alpha_{g_i}$$.

h is not a homomorphism, correct?

5. Nov 30, 2007

### ehrenfest

does anyone understand this question?

6. Nov 30, 2007

### quasar987

The group of alpha_g? I assume you mean the subgroup of $$S_G$$ obtained from the map h you defined lastly?

If so, then that's just Cayley's theorem.

In Cayley's theorem, we show that every group G is isomorphic to a subgroup of $$S_G$$ (the group of permutation of G.)

To do this, we show that the map $$h:G\rightarrow S_G$$ defined as you did is a monomorphism. Then the restriction of its codomain to h(G) is an isomorphism.

7. Nov 30, 2007

### ehrenfest

The group I speak of is $$\{\alpha_{g} : g \in G\}$$ under composition of functions.

I didn't realize that but you're right. So my question really is then, to prove Cayley's Theorem can you define $$\alpha_{g}(g_i) = g_i*g$$ or do you need to define
$$\alpha_{g}(g_i) = g*g_i$$?

I think that we need the latter, but I'm afraid I am making a stupid mistake.

8. Nov 30, 2007

### quasar987

IT works either way. As you said, it's "symetric"

1° For all g, $$\alpha_{g}(g_i) = g_i*g$$ defines a permutation of G, because 1) if g1*g=g2*g, then discover that g1=g2 by multiplying by g^-1 from the right. That's injectivity. 2) let g' be in G, then $$g_i=g'*g^{-1}$$ is sent to g' by $$\alpha_{g}$$. That's surjectivity.

2° Show h is an homomorphism. That's routine.

3° Let g be in Ker(h). Then $$\alpha_{g}=id_G$$. So for any g' in G, g'*g=g'. This can only be so if g=1. Hence the kernel is trivial, hence h is a monomorphism.

9. Nov 30, 2007

### quasar987

Funny!

2° $$h(g*g')(g_i)=\alpha_{g*g'}(g_i)=g_i*(g*g')=(g_i*g)*g'=\alpha_{g'}(g_i*g)=\alpha_{g'}(\alpha_{g}(g_i))=(\alpha_{g'}\circ \alpha_{g})(g_i)=[h(g')\circ h(g)](g_i)$$

so it seems you were right in worrying! It's not a morphism if we multiply from the right!

10. Nov 30, 2007

### ehrenfest

I claim that h is not a homomorphism in this case:

$$\alpha_{g_1*g_2}(g_i) = g_i*g_1*g_2$$
but
$$\alpha_{g_1}\circ \alpha_{g_2}(g_i) = \alpha_{g_1}(\alpha_{g_2}(g_i)) = g_i*g_2*g_1$$

EDIT: you beat me

Last edited: Nov 30, 2007
11. Nov 30, 2007

### quasar987

It works if we define $$\alpha_g$$ to act on gi by multiplication from the right by the inverse of g instead:

$$\alpha_g(g_i)=g_ig^{-1}$$.

Because now

$$\alpha_{g*g'}(g_i)=g_i(g*g')^{-1}=g_ig'^{-1}g^{-1}=(\alpha_{g}\circ\alpha_{g'})(g_i)$$.

Last edited: Nov 30, 2007
12. Nov 30, 2007

### quasar987

Yes, morphism and homomorphism are interchangeable terms.

13. Nov 30, 2007

### ehrenfest

I just realized that and deleted the post. But thanks. Everything is clear now.