Homomorphism - Sharper Cayley Theorem

[Kiwi1]

I am asked:

Prove that $$G \rightarrow S_X$$ defined by $$h(x) = \rho _a$$ is a homomorphism.

So I must prove that for any $$a,b \in G$$ h(a)h(b) = h(ab).

But must I also prove seperately that: h is ONTO $$S_X$$?

 

[Euge]

Hi Kiwi,

There are pieces of information that you've left out in your prompt. What is $X$ and $\rho_a$? When you write $h(x) = \rho_a$, what is the relationship between $x$ and $a$?

 

[Kiwi1]

Hi Kiwi,

There are pieces of information that you've left out in your prompt. What is $X$ and $\rho_a$? When you write $h(x) = \rho_a$, what is the relationship between $x$ and $a$?

Hi Euge

I really just wanted to know in general, if I am asked to prove a function is a homomorphism must I prove that it is onto?

In any case I have attached an image of the question from my text. I have done part 1.

I think the answer must be no because for this particular problem:
Let $G = \{e, a, a^2 ... a^{3k-1}\}$, and
Let $H = \{e, a^3, a^6 ... a^{3k-3}\}$, then

$X = \{H, aH, a^2H\}$, and there is no $\rho _a$ that fixes H and exchanges the other two cosets. Therefore there is a member of $S_3$ that cannot be reached by any $\rho_a$.

Question 4 also seems to give me a clue that h is not expected to be onto?

 

[Euge]

If $(G,\cdot)$ and $(G',*)$ are groups, a function $f : (G,\cdot) \to (G',*)$ is a group homomorphism if $f(a\cdot b) = f(a) * f(b)$ for all $a,b\in G$. Note that the definition does not require $f$ to be onto. In the case of your problem, to prove $h$ is a homomorphism, you must show $\rho_a\rho_b = \rho_{ab}$ for all $a,b\in G$.