# Guided proof to the isomorphism theorems.

1. Apr 4, 2007

### *melinda*

1. The problem statement, all variables and given/known data
Let $G_1$ and $G_2$ be groups with normal subgroups $H_1$ and $H_2$, respectively. Further, we let $\iota_1 : H_1 \rightarrow G_1$ and $\iota_2 : H_2 \rightarrow G_2$ be the injection homomorphisms, and $\nu_1 : G_1 \rightarrow G_1/H_1$ and $\nu_2 : G_2/H_2$ be the quotient epimorphisms.

Given that there exists a homomorphism $\sigma : G_1 \rightarrow G_2$, show that there exists a unique mapping $\overline{\sigma} : G_1/H_1 \rightarrow G_2/H_2$ such that $\overline{\sigma} \circ \nu_1 = \nu_2 \circ \sigma$ if and only if $\sigma[H_1] \subset H_2$. If such a $\overbar{\sigma}$ exists, it is a homomorphism.

2. Relevant equations

There aren't any equations, as this is a proof.

3. The attempt at a solution

I know that since $\nu_1$ and $\nu_2$ are epimorphisms, they are surjective homomorphisms. So $Im(\nu_1)=G_1/H_1$ and $Im(\nu_2)=G_2/H_2$. But I really don't see how to get this proof off the ground. Please help get me started.

The next question reads as follows.

Prove that there exists a unique mapping $\sigma^{\prime} : H_1 \rightarrow H_2$ such that $\iota_2 \circ \sigma^{\prime} = \sigma \circ \iota_1$ if and only if $\sigma[H_1] \subset H_2$. If such a $\sigma^{\prime}$ exists, it is a homomorphism.

Last edited: Apr 4, 2007
2. Apr 4, 2007

### matt grime

Let s=sigma, v=v_1 and w=v_2, cos I want to do this without having to type in tex. The only place to start is with the composite ws, since that is a well defined map, and it goes from G_1 to G_2/H_2. This gives a map of G_1/H_1 if and only if H_1 is in the kernel of the map ws. Which is if and only if.... That is existence. Uniqueness we'll come to in a second.