# In each of the following cases, determine whether or not $G$ is isomorphic to the product group $H \times K$.

In each of the following cases, determine whether or not $$G$$ is isomorphic to the product group $$H \times K$$.

a) $$G$$= {invertible upper triangular $$2\times2$$ matrix },$$H=$${invertible diagonal matrices} $$K$$={upper triangular matrices with diagonal entries $$1$$}

b)$$G = \mathbb{C}^{\times}$$, $$H$$ = {unit circle} , $$K$$= {postive real number}

# My attempt : in both a) and b) cases $$G$$ is not isomorphic to the product group $$H \times K$$.

i was thinking about first theorem of isomorphism

For $$a)$$ $$\frac{G}{H}$$ not isomorphics to $$K$$, because H is not normal subgroup of $$G$$, so $$G \neq H \times K$$

for $$b)$$ $$\frac{G}{K} =\frac{\mathbb{C}^{\times}}{\text{postive real number}}$$will isomorphic to $$\{ -1,+1\}$$, that is it $$\{-1,+1\}$$ $$\neq$$ unit circle, so $$G \neq H \times K$$

Is it true ??

any hints/solution will be appreciated

thanks u

$$a)$$ is False ,take
$$\begin{bmatrix} 2 &0 \\ 0& 1 \end{bmatrix}\begin{bmatrix} 1 &2 \\ 0& 1 \end{bmatrix} \neq \begin{bmatrix} 1 &2 \\ 0& 1 \end{bmatrix}\begin{bmatrix} 2 &0 \\ 0& 1 \end{bmatrix}$$.
$$b)$$ is True , since $$\mathbb{C^×}$$ is abelian, both K and H are normal subgroups. Moreover the two subgroups intersect only in the identity of $$\mathbb{C}^{\times}$$.The product $$KH$$ is the whole $$\mathbb{C}$$ follows from the fact that every complex number has an expression in polar coordinates, that is as a product of a positive real number and a number in the unit circle