= Solution
A <subgroup> $H$ of a <group> $G$ is a <normal subgroup> when $gHg^{-1}=H$ 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> $\varphi:G\to K$, its <kernel of a group homomorphism> is normal and
$$
G/\ker\varphi\cong\operatorname{im}\varphi,
\qquad g\ker\varphi\longmapsto\varphi(g).
$$
The 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:
$$
\varphi\!\begin{pmatrix}a&b\\0&d\end{pmatrix}=(a,d)
\quad\text{in}\quad\mathbb R^\times\times\mathbb R^\times.
$$
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 $(aa',dd')$, proving the <group homomorphism> property. Any pair of nonzero diagonal entries is attained by a <diagonal matrix>, so $\varphi$ is surjective. Its <kernel of a group homomorphism> consists precisely of the <matrices> with $a=d=1$, namely $H$. Hence
$$
\boxed{H\trianglelefteq G,\qquad G/H\cong(\mathbb R^\times)^2}
$$
with coordinatewise multiplication, not addition on $\mathbb R^2$.
Back to article page
