1
$\begingroup$

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

$\endgroup$
3
$\begingroup$

$a)$ 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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.