1
$\begingroup$

I have a subgroup of $GL(3, \mathbb{R})$, given by

\begin{pmatrix} a & 0 & b \\ 0 & a & c\\ 0 & 0 & d \end{pmatrix}

with a,d different from 0 and a,b,c and d real numbers. Let's call this subgroup G.

Now, we consider the following subgroups of G:

\begin{pmatrix} 1 & 0 & b \\ 0 & 1 & c\\ 0 & 0 & 1 \end{pmatrix}

\begin{pmatrix} a & 0 & 0 \\ 0 & a & 0\\ 0 & 0 & d \end{pmatrix}

Let's call N the first one and Q the second one. Now, I've proven that N is a normal subgroup of G, but now I have to show that G/N is isomorphic to Q and decide if G is isomorphic to $N \times Q$. I suppose that I have to use the isomorphism theorems for groups. Since I have N normal subgroup and Q, then I've been trying to use the second one with no success.

$\endgroup$
  • $\begingroup$ Hint For any $\pmatrix{a&0&b\\0&a&c\\0&0&d} \in G$ and $n \in N$, we have $gn = \pmatrix{a&*&*\\0&a&*\\0&0&d}$ for some entries $*$. $\endgroup$ – Travis Willse Jan 19 '17 at 12:05
  • $\begingroup$ N, Q are abelian $\endgroup$ – i. m. soloveichik Jan 22 '17 at 14:56
0
$\begingroup$

For technical reasons I will represent the elements of $N$ as $\begin{pmatrix} 1 & 0 & x\\0 & 1 & y\\0 & 0 & 1\end{pmatrix}$ and the elements of $Q$ as $\begin{pmatrix}u & 0 & 0\\0 & u & 0\\0 & 0 & v\end{pmatrix}$. If we take an arbitrary element $g \in G$ defined by some $a, b, c$ and $d$ and consider its left coset $gn$, where $n$ runs through $N$ we see that these elements are of the form \begin{pmatrix}a & 0 & a x + b\\0 & a & a y + c\\0 & 0 & d\end{pmatrix}. For the particular choice of $x = -b/a$ and $y = -c/a$ we see that this element belongs to $Q$ with $u = a$ and $v = a$. We have shown that $\overline{g} \cap Q = \{q\}$. In other words, we can take for every element of the quotient (a left coset) a representative in $Q$. This shows that $G = N \rtimes Q$, the semidirect product of $N$ and $Q$.

| cite | improve this answer | |
$\endgroup$

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.