prove $G \cong H \times N$, $G= \{ a_{ij} \in GL_n(F) :a_{ij} =0 \text{ if } i > j \text{ and } a_{11} =a_{22} = \dots =a_{nn}\}$

Question

Let $$G= \{ a_{ij} \in GL_n(F) :a_{ij} =0 \text{ if } i > j \text{ and } a_{11} =a_{22} = \dots =a_{nn}\}$$

where $F$ is a field, (in words) the group of upper triangular matrices all whose diagonal are equal

Prove $$ G= D \times U $$

where $D$ is a group of nonzero multiples of the identity matrix

$U$ is a group of upper triangular with 1's in diagonal

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This is from Dummit's ch 5.3 on second edition,

Thm 9: $G$ is a group with subgroup $H$ and $K$

s.t

1 $H$ and $K$ are normal in $G$

2 $H \cap K =\{e\}$

3 $HK \cong H \times K$

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1.a need to show that $D $ normal in $G$?

that is $g d g^{-1} \in D$

has to do with the determinant

$det(gdg^{-1})=det(g) det(g)^{-1} det(d) = det(d)$

so $d\in D$

1.b U is normal in $G$

$g u g^{-1} \in U$

2. $D \cap U =\{ e \}$

obvious? (or is it a whole new problem of its own)

3. $HN=G=H \times N$

(might want to make this a new question)

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Is this somewhere in the ballpark? is this the correct approach? better?

Answer 1

1. While the working is accurate, it doesn't prove that $D$ is normal in $G$. $det(A) = det(B) \not\implies A$ has the same form as $B$ in general.

You could justify it by showing a direct calculation with an arbitrary diagonal matrix in this case, but that justification can also be used more directly.

2. Obvious, but could use a sentence or two of justification ("Clearly, the only matrix that satisfies both predicates has 1s in the diagonal...")

3. By noting that multiplication by a scalar matrix (i.e. an element of $D$) scales up each element of the matrix by a fixed amount, you should be able to show that $HN = G$.

For $g \in G$ there exists a scalar $s$ such that $sg$ has only 1s in the diagonal. So you there exists a corresponding map $s^{-1}u$ to $g$, for all $g \in G$

However as Yeldarbskich suggested, it might be easier to directly show an isomorphism than invoking the theorem.