Let $\mathbb{F}$ be a field. Consider the following three groups-
$$G=\left\{\begin {pmatrix} 1&a&b\\ 0&1&c\\ 0&0&1\\ \end{pmatrix} : a,b,c\in \mathbb{F}\right\} $$
$$H=\left\{\begin {pmatrix} 1&0&d\\ 0&1&0\\ 0&0&1\\ \end{pmatrix} : d\in \mathbb{F}\right\} $$
$$T=\left\{\begin {pmatrix} x\\ y \end{pmatrix} : x,y\in \mathbb{F}\right\} $$
Prove that $G/H \cong T$.
First I showed that $H=Z(G)$. Then I concluded that $G/H=G/Z(G)\cong Inn(G)$.
I thought I might construct an isomorphism between $T$ and $Inn(G)$ and then use transitivity of isomorphism.
I calculated what a function $\phi$ in $Inn(G)$ would look like, and got: $\phi \left( \begin {pmatrix} 1&a&b\\ 0&1&c\\ 0&0&1\\ \end{pmatrix} \right)= \begin {pmatrix} 1&\tilde a&\tilde b\\ 0&1&\tilde c\\ 0&0&1\\ \end{pmatrix} \times \begin {pmatrix} 1&a&b\\ 0&1&c\\ 0&0&1\\ \end{pmatrix} \times \begin {pmatrix} 1&\tilde a&\tilde b\\ 0&1&\tilde c\\ 0&0&1\\ \end{pmatrix} ^{-1} = \begin {pmatrix} 1& a& b-a\tilde c+ c\tilde a\\ 0&1&c\\ 0&0&1\\ \end{pmatrix}$
So the result is depended only on $\tilde a$ and $\tilde c$ and not on $\tilde b$.
"Well, that's great!", I thought to myself. "I can now construct $\psi : Inn(G) \rightarrow T$ to be $\psi(\phi_g)= \begin {pmatrix} \tilde a\\ \tilde c\\ \end{pmatrix} $ when $g= \begin {pmatrix} 1&\tilde a&\tilde b\\ 0&1&\tilde c\\ 0&0&1\\ \end{pmatrix} $"
But unfortunately, I wasn't able to prove that $\psi$ is an isomorphism (it is not injective).
Then I came here and spent an hour learning lots of $\LaTeX$in order to write this.
Can anyone get me out of this conundrum? Many thanks!
Do the multiplication $$ \begin{pmatrix} 1&a&b\\ 0&1&c\\ 0&0&1 \end{pmatrix} \begin{pmatrix} 1&a'&b'\\ 0&1&c'\\ 0&0&1 \end{pmatrix} =\begin{pmatrix} 1&a'+a&b'+ac'+b\\ 0&1&c'+c\\ 0&0&1 \end{pmatrix} $$ which means that the map $$ f\colon G\to T,\qquad \begin{pmatrix} 1&a&b\\ 0&1&c\\ 0&0&1 \end{pmatrix}\mapsto\begin{pmatrix}a\\c\end{pmatrix} $$ is a homomorphism. What's the kernel of $f$?