In general, how does one show that a mapping between two groups is a monomorphism? Example included.

Question

The definition of a group monomorphism is as follows:

Let $G$ and $H$ be groups and suppose $\phi:G\to H$ is a group homomorphism. Then $\phi$ is a group monomorphism if and only if $\phi$ is an injection.

There are so many morphisms to keep track of in group theory that I some times lose track of what I am trying to prove. If we go by this definition, it suffices to show that $\phi$ is a well-defined homomorphism between $G$ and $H$ and then to show that $\phi$ is injective.

So here is an example from Algebra (Hungerford) page 64:

Let $\left\{G_i | i\in I\right\}$ be a family of groups and $J\subset I$. The map $\alpha: \prod_{j\in J} G_{j} \to \prod_{i\in I} G_{i}$ given by $\left\{a_{i}\right\}\mapsto \left\{b_{i}\right\}$, where $b_{j} = a_{j}$ for $j\in J$ and $b_{i}=e_{i}$ for $i \notin J$, is a monomorphism of groups and $\prod_{i\in I} G_{i} /\alpha(\prod_{j\in J} G_{j})$ is isomorphic to $\prod_{i\in I-J} G_{i}$.

Based on my understanding, the first part of this is to show that $\alpha$ is well-defined and then to show it is injective. Am I understanding this correctly?

The second part showing $\prod_{i\in I} G_{i} /\alpha(\prod_{j\in J} G_{j})$ is isomorphic to $\prod_{i\in I-J} G_{i}$ is not really a concern at the moment.

Answer 1

Yes. You need to show that $\alpha$ is a well-defined group homomorphism then show that it's injective.

Answer 2

My attempt at proving $\alpha$ is an injective homomorphism:

By hypothesis the mapping $\alpha$ is a well-defined function since $a_j = b_j$ for all $j\in J$. We must now show that $\alpha$ is an injective homomorphism. Define $H = \prod_{j\in J} G_{j}$ and $K = \prod_{i\in I} G_{i}$ and let $\left\{a_j\right\}$, $\left\{c_j\right\} \in H$ and assume $\alpha(\left\{a_j\right\}) = \alpha(\left\{c_j\right\})$. Then we obtain

$$\alpha(\left\{a_j\right\}) = \left\{b_j\right\}$$

and

$$\alpha(\left\{c_j\right\}) = \left\{b_j\right\}$$

for all $j\in J$. By hypothesis, we know that $b_j = c_j$ and $b_j = c_j$ for all $j\in J$. Then $a_j = c_j$ by the transitive property. So for all $j\in J$, the function $\alpha$ is injective. For any $i\notin J$, the function $\alpha$ maps all elements of $H$ to the identity elements $e_i \in G_i$.

Now choose any two elements $\left\{a_j\right\}$, $\left\{c_j\right\} \in H$, then compute $\alpha(\left\{a_j\right\}\left\{c_j\right\})$. We obtain

$$\alpha((a_j, c_j))$$

but from above we have $a_j = b_j$ and $c_j = b_j$ for all $i\in J$. Then

$$\alpha((a_j,c_j)) = (b_j, b_j) = \left\{b_j\right\}\times\left\{b_j\right\}$$

This is simply $\alpha(\left\{a_j\right\})\cdot \alpha(\left\{c_j\right\})$. So $\alpha$ is an injective homomorphism.

I'm not sure about the homomorphism part because I'm not accustomed to working with sets like these. Please let me know if I am way off base or if my proof just needs some adjustments (Note: I am not proving the homomorphism part yet).