 Solution

A subgroup $H$ of a group $G$ is a normal subgroup when $gH{g}^{-1}=H$ for every $g\in G$, equivalently $gH=Hg$. The quotient group $G/H$ has the left cosets as elements, with $(gH)(kH)=gkH$, identity $H$ and inverse $(gH{)}^{-1}=g^{-1}H$. Normality makes this multiplication independent of representatives.

The first isomorphism theorem for groups says that for a group homomorphism $\phi :G\to K$, its kernel of a group homomorphism is normal andThe image is taken with the induced group operation; surjectivity onto all of $K$ is not required.

For the given upper triangular matrices, take diagonal entries:The target is a direct product of groups, with multiplication of nonzero real numbers in each coordinate. Indeed the product of two such matrices has diagonal $(a{a}^{\prime },d{d}^{\prime })$, proving the group homomorphism property. Any pair of nonzero diagonal entries is attained by a diagonal matrix, so $\phi$ is surjective. Its kernel of a group homomorphism consists precisely of the matrices with $a=d=1$, namely $H$. Hencewith coordinatewise multiplication, not addition on ${\mathbb{R}}^2$.