Prove $D \cong (F,+)$, the additive group in $F$.

Question

Prove $D \cong (F,+)$, with

$D: \{ \begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix} \mid b \in F \}$

Can somebody help me out as to showing these two groups are isomorphic? Thanks.

Answer 1

The obvious bijection from $D$ to $F$ maps $\pmatrix{1&b\\0&1}$ to $b$.

Furthermore, it's a homomorphism because $\pmatrix{1&b\\0&1}\pmatrix{1&c\\0&1}=\pmatrix{1&b+c\\0&1}$.

Therefore, this map is an isomorphism.

Answer 2

One of the main tools to show an isomorphism is the homomorphism theorem (which we do not really use here.....)

For that we want to find a surjective homomorphism with trivial kernel

$f:D\to F.$

It makes sense to define the map by $\begin{pmatrix}1&b\\0&1\end{pmatrix}\mapsto b$, as we want to relate matrices with elements of the field, and by definition of $D$ this is kind of the only sensible way to do so.

Now we have to show that this is indeed a homomorphism. That the function is surjective, and has trivial kernel (Which implies that $f$ is injective. Why?). This shows that we have an isomorphism.

I leave the details up to you. But don't worry; this is easy to check. Or feel free to ask.