Let $\mathbb H$ be the set of all $3\times 3$ matrices of the form $\begin{pmatrix}1 & a & b\\0 & 1 & c\\0 & 0 &1\\ \end{pmatrix}$ where $a,b,c\in \mathbb R$ .I have to show that it is a Lie group.I have thought of doing it in the following way:
My Approach
First of all it is evident that $\mathbb H$ is a subgroup of $GL(3,\mathbb R)$.It is easy to show.Now I claim that $\mathbb H$ is a closed subset of $GL(3,\mathbb R)$ with the usual Euclidean topology.Since there is a norm in $M(3,\mathbb R)$ so we can use that.Now if $\{H_n\}$ is a sequence in $\mathbb H$ which converges to $H\in M_3(\mathbb R )$ in the Euclidean norm,then its upper triangular coordinates which are $a_n,b_n,c_n$ (say) converge in $\mathbb R$ to $a,b,c$ and so by componentwise continuity $H_n\to H$ as $n\to \infty$.
Thus by sequential criterion for closed sets,saying any convergent sequence in $\mathbb H$ has its limit in the set $\mathbb H$,we can conclude $\mathbb H$ is closed.Now a closed subgroup of a Lie group is a Lie group.So,we are through.
This was my proof and I think this one is the most easy one.Can someone tell me if I have done it correctly or there is some error in the above proof?