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Assume all matrices are $n \times n$ matrices. If $G$ is the general linear group (the set of all invertible matrices with entries in $\mathbb{R}$), $K$ is the special linear group (the set of all matrices of determinant 1, a subgroup of $G$), and $H = \{aI: a \neq 0\}$ (where $I$ is the $n \times n$ identity matrix), show that $HK = G$.

I am using set containment to prove this statement. It is easy to show $HK \subseteq G$. Note that I showed that $HK = \{ak: k \in K\text{, } a \neq 0\}$.

Now if $X \in G$, $X$ is invertible, therefore making $\det(X) \neq 0$. So I need to show that there exists a $k \in K$ and $a \neq 0$ such that $X = ak$, where $\det(k) = 1$. I'm not sure how to do this part.

Edit: $n$ is odd.

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  • $\begingroup$ it seems that if $n$ is even then there is no way to get $\det <0$ matrices from $HK$. $\endgroup$ – janmarqz Jan 30 '14 at 2:59
  • $\begingroup$ Sorry, $n$ is odd. Forgot about that assumption. $\endgroup$ – Clarinetist Jan 30 '14 at 3:07
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Here is a generalization: for any group $G$, homomorphism $f:G\to H$, kernel $\ker f=K$, and subset $X\subseteq G$, we necessarily have $f(X)=H~\Rightarrow~G=XK$.

Can you prove this? To do so, you want to be able to write an arbitrary $g\in G$ as $xk$ with $x\in X$ and $k\in K$. (To find an $x$ to use, try applying $f$ to $g=xk$ and thinking about $f(X)=f(G)$.)

Do you see how this applies to your case? There is a group, a homomorphism (?), a kernel (?)...

(Note how important it is that every real number is an $n$th power; this is why $n$ must be odd.)

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Hint:

Suppose M has nonzero determinant $d$. Since n is odd, $d^{1/n}=a$ exists.

What is the determinant of $aI\cdot M$ ?

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